The sum of the a(n) first digits of S is a prime
Hello
SeqFans  here is another nightmare:
«
Smallest integer not yet present in S
such that the sum of the a(n) first digits of S is a prime »:
I
get by hand:
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,34,49,37,200,36,38,39,42,59,51,54,56,61,69,201,...
Sum
of the first 2 digits is a prime: 2+1 = 3
Sum
of the first 1 digit is a prime: 2 = 2
Sum
of the first 4 digits is a prime: 2+1+4+6 =
13
Sum
of the first 6 digits is a prime:
2+1+4+6+3+7= 23
Sum
of the first 3 digits is a prime: 2+1+4 = 7
Sum
of the first 7 digits is a prime:
2+1+4+6+3+7+8 = 31
Sum
of the first 8 digits is a prime:
2+1+4+6+3+7+8+6 = 37
Sum
of the first 60 digits is a prime: 2+1+4+6+3+7+8+6+0+9+1+1+1+4+1+7+6+1+1+6+2+2+2+5+3+0+2+6+2+8+3+4+4+9+3+7+2+0+0+
3+6+3+8+3+9+4+2+5+9+5+1+5+4+5+6+6+1+6+9+2
= 241
Sum
of the first 9 digits is a prime:
2+1+4+6+3+7+8+6+0 = 37
Sum
of the first 11 digits is a prime:
2+1+4+6+3+7+8+6+0+9+1 = 47
Sum
of the first 14 digits is a prime:
2+1+4+6+3+7+8+6+0+9+1+1+1+4 = 53
Sum
of the first 17 digits is a prime:
2+1+4+6+3+7+8+6+0+9+1+1+1+4+1+7+6 = 67
Sum
of the first 61 digits is a prime: 2+1+4+6+3+7+8+6+0+9+1+1+1+4+1+7+6+1+1+6+2+2+2+5+3+0+2+6+2+8+3+4+4+9+3+7+2+0+0+
3+6+3+8+3+9+4+2+5+9+5+1+5+4+5+6+6+1+6+9+2+0
= 241
Sum
of the first 16 digits is a prime:
2+1+4+6+3+7+8+6+0+9+1+1+1+4+1+7 = 61
... 

>Above
column is S
In
computing this, one is forced to put constraints on the future of S. See here,
for instance:
S = 2,1,4,...
As
we cannot write "3" after "1" (because the sum of the first
"3" digits would be 6  a composite), we write "4" 
leaving the future "open". But this "4" forces the next
term, 6:
S = 2,1,4,6,...
And
again, this "6" puts a constraint on the future of S. "3"
is forced, as "3" is the « smallest integer not yet present in S such
that the sum of the a(n) first digits of S is a prime »; indeed, it works,
"3" describes the "past" of S and 2+1+4=7, a prime.
S = 2,1,4,6,3,7,...
"7"
resolves the constraint put by "6"; indeed the sum of the first 6 digits
of S is a prime  "7" being the smallest available integer: 2+1+4+6+3+7=
23
This
"7" puts another constraint on the future of S  resolved by
"8":
S = 2,1,4,6,3,7,8,... Check:
2+1+4+6+3+7+8 = 31, ok.
This
"8", again, puts a constraint on the next term  and we notice quickly
that we must take "60" as the next term! If we could have taken "6"
instead, we would have had a prime sum (31+6=37)  but 6 was already in S! We
must try something else  and after having checked that no smallest integer
fits, we write "60" after "8":
S = 2,1,4,6,3,7,8,60,...
This
puts a huge constraint on S! We have to remember that the sum of the first 60
digits of S is a prime!
Nevertheless,
we proceed... (fast forward until):
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,...
The
above last digit ("8") is the 30st of S;
the digit sum so far is 107;
What
would be the next term "xy", after 28? All
"past" prime sums have been described so far  thus we write
"for the future" and try "31":
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,31,...
No
 the "3" of "31" is the 31st digit of S; as
107+"3" is 110 (composite) we cannot keep "31".
We try "32":
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,32,...
No
again: the sum of the 32 first digits would now be
107+"3"+"2" = 112, a composite.
"33"
seems ok  as the sum of the 33 first digits is prime
(read the "n^{th} digit" vertically and the cumulative sums
"CS" vertically too):
n^{th} = 1 2 3 4 5 6 7 89 1 11 11 11
11
12 22 22 22
22
23 33 33
n^{th} 0 12 34 56 78 90 12 34 56 78 90 12 34
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,33,xy
CS= 2 3 7 1 1 2 3 33 4 44 45 56
66 67 77 88 88 99 91 11 .
CS 3 6 3 1 77 6 78 93 41 78 95 79 16 99 17 90 11 .
7 03 .
Ouch...
We have a problem with the next digit, "x"! No digit will fit, as the
closest prime to 113 is 127  14 units apart! We see that 33 puts too strong a constraint on the future of S  a halting
one! We thus erase "33" and try "34" (shouting ‘alleluia’
here to the inventor of the copy/paste technique):
n^{th} =1 2 3 4 5 6 7 89 1
11 11
11
11
12 22 22 22
22
23 33 33
n^{th} 0 12
34 56 78 90 12 34 56 78 90 12 34
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,34,xy ...
CS= 2 3 7 1 1 2 3 33 4 44 45 56
66 67 77 88 88 99 91 11 .
CS 3 6 3 1 77 6 78 93 41 78 95 79 16 99 17 90 11 .
7 04 .
Now,
with "34", we have 2 available digits to bridge the gap; the smallest
integer fitting will be 49:
n^{th}= 1 2 3 4 5 6 7 89 1 11 11 11 11 12 22 22 22
22
23 33 33
n^{th} 0 12
34 56 78 90 12 34 56 78 90 12 34
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,34,49, ...
CS= 2 3 7 1 1 2 3 33 4 44 45 56
66 67 77 88 88 99 91 11 11
CS 3 6 3 1 77 6 78 93 41 78 95 79 16 99 17 90 11 12
7 04 87
Etc.
I
don’t now if such a seq can exist; there might be for
ever the doubt that we have to erase a lot of terms in order to proceed  and
"a lot of terms" might force us, at some (far away) point, NOT to
start with S = 2,1,4,6,3,7,8,60,...
Best,
É.
__________
Maximilian Hasler was quick to submit this:
[Éric] :
>
S =
2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,34,49,37,200,36,38,39,42,59,51,54,56,61,69,201,...
[Maximilian]:
not bad... but I get something different
for the last 3 of your terms, when I compute some more :
The
following 86 terms satisfy the criteria ("so far"):
[2,
1, 4, 6, 3, 7, 8, 60, 9, 11, 14, 17, 61, 16, 22, 25, 30, 26, 28, 34, 49, 37,
200, 36, 38, 39, 42, 59, 51, 54, 56, 62, 71, 600, 65, 70, 66, 75, 201, 72, 67,
68, 82, 100, 83, 84, 91, 93, 300, 94, 95, 301, 98, 101, 103, 110, 116, 120,
302, 121, 130, 132, 202, 133, 141, 500, 142, 143, 148, 150, 154, 155, 165, 303,
166, 174, 203, 175, 183, 204, 184, 192, 196, 290, 208, 800, ...]
__________
Douglas McNeil
answered:
[Éric] :
>>
S = 2,1,4,6,3,7,8,60,9,11,14,17,61,16,22,25,30,26,28,34,49,37,200,36,38,39,42,59,51,54,56,61,69,201,...
[Doug]:
>
but this has 61 twice
[Éric] :
... aaargghllll!
[Doug]:
M. Hasler wrote:
>
[2, 1, 4, 6, 3, 7, 8, 60, 9, 11, 14, 17, 61, 16, 22, 25, 30, 26, 28, 34, 49,
37, 200, 36, 38, 39, 42, 59, 51, 54, 56, 62, 71, 600, 65, 70, 66, 75, 201, 72,
67, 68, 82, 100, 83, 84, 91, 93, 300, 94, 95, 301, 98, 101, 103, 110, 116, 120,
302, 121, 130, 132, 202, 133, 141, 500, 142, 143, 148, 150, 154, 155, 165, 303,
166, 174, 203, 175, 183, 204, 184, 192, 196, 290, 208, 800, ...]
... which seems to selfdescribe so far, but I find
sage: S[:100]
[2,
1, 4, 6, 3, 7, 8, 60, 9, 11, 14, 17, 61, 16, 22, 25, 30, 26, 28, 34, 49, 37,
200, 36, 38, 39, 42, 59, 51, 54, 56, 62, 68, 2000, 63, 67, 80, 69, 70, 72, 73, 82, 90,
600, 81, 83, 91, 601, 84, 92, 97, 103, 201, 104, 106, 107, 112, 113, 120, 126,
130, 202, 131, 139, 400, 138, 140, 141, 144, 148, 153, 163, 203, 164, 168, 172,
178, 181, 900, 182, 183, 187, 196, 500, 280, 800, 197, 198, 204, 206, 208, 224,
290, 225, 232, 602, 233, 234, 238, 246]
which I think also selfdescribes but
disagrees from the 33rd term, which I take as 68. I didn’t use a search limit,
which probably explains the difference as the next term seems to be 2000.
But
who knows? If the above is right (even setting aside the possibility of
backwardpropagating blockers later, which I’m too sleepy to think about) it’d
be the only time I’d ever successfully computed one of Eric’s selfreferencing
sequences in any of the first halfdozen attempts..
Doug
__________
Maximilian:
I
confirm Doug’s terms:
By
setting my search limit to 3000, I get the same first 100 terms (How come I did
not think of trying this??), and they remain stable when I calculate 200 of
them.
But
the fact that my terms were consistent to over 85 terms but wrong from the 33rd
one on, makes it clear how little we can trust in these results...
(Which
does not mean that the question of existence of the sequence is hopeless to
answer, IMO).
Maximilian
try2complete(100,,2001)
[2, 1, 4, 6, 3, 7, 8, 60, 9, 11, 14, 17, 61, 16, 22, 25, 30, 26, 28, 34,
49, 37, 200, 36, 38, 39, 42, 59, 51, 54, 56, 62, 68, 2000, 63, 67, 80, 69, 70,
72, 73, 82, 90, 600, 81, 83, 91, 601, 84, 92, 97, 103, 201, 104, 106, 107, 112,
113, 120, 126, 130, 202, 131, 139, 400, 138, 140, 141, 144, 148, 153, 163, 203,
164, 168, 172, 178, 181, 900, 182, 183, 187, 196, 500, 280, 800, 197, 198, 204,
206, 208, 224, 290, 225, 232, 602, 233, 234, 238, 246]
try2complete(200,,2001) =
[2, 1, 4, 6, 3, 7, 8, 60, 9, 11, 14, 17, 61, 16, 22, 25,
30, 26, 28, 34, 49, 37, 200, 36, 38, 39, 42, 59, 51, 54, 56, 62, 68, 2000, 63,
67, 80, 69, 70, 72, 73, 82, 90, 600, 81, 83, 91, 601, 84, 92, 97, 103, 201, 104,
106, 107, 112, 113, 120, 126, 130, 202, 131, 139, 400, 138, 140, 141, 144, 148,
153, 163, 203, 164, 168, 172, 178, 181, 900, 182, 183, 187, 196, 500, 280, 800,
197, 198, 204, 206, 208, 224, 290, 225, 232, 602, 233, 234, 238, 246, 248, 253,
401, 252, 254, 265, 402, 266, 269, 271, 278, 403, 281, 285, 340, 291, 292, 294,
295, 301, 309, 312, 317, 360, 318, 321, 328, 329, 333, 337, 341, 901, 346, 801,
347, 350, 355, 369, 357, 365, 373, 410, 378, 379, 603, 380, 382, 383, 384, 393,
404, 502, 604, 409, 501, 413, 418, 802, 419, 424, 428, 433, 605, 434, 437, 438,
441, 442, 447, 457, 503, 458, 464, 469, 1000, 470, 471, 472, 485, 902, 486, 488,
493, 499, 606, 607, 504, 510, 511, 513, 515, 521, 526, 531, 690, 532, 538, 542,
700, 543]
__________
We
now have 200 stable terms: many thanks to both of you, Maximilian and Doug!
Best,
É.
[Augustus
31^{st}, 2010]