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%I #13 Jul 12 2014 18:01:23
%S 11142,11553,14088,16713,18801,22284,23097,23718,26787,28818,323589,
%T 327939,328992,416103,438357,459069,502149,595194,617928,647178,
%U 656457,665853,677019,682230,747099,767748,775782,799233,813861,832986,847266,855897,858648,862014,924366,970767,10174023,10240146
%N Numbers x such that the base 10 representation of x^2 forms an arithmetic sequence when split into equal-sized chunks.
%C This sequence only includes numbers which produce arithmetic sequences of at least three terms (and in fact, no squares containing sequences of more than three terms have been found).
%H Michel Marcus, <a href="/A244660/b244660.txt">Table of n, a(n) for n = 1..142 (up to 10^8)</a>
%e 11142^2 = 124144164 and 124,144,164 is an arithmetic sequence.
%o (Python)
%o from itertools import count
%o .
%o def check(n,power):
%o ...np = str(n**power)
%o ...l=len(np)
%o ...for chunks in range(3,5):
%o ......if l%chunks==0:
%o .........step = l//chunks
%o .........bits = [int(np[i:i+step]) for i in range(0,l,step)]
%o .........diff = bits[1]-bits[0]
%o .........old=bits[1]
%o .........go = True
%o .........for bit in bits[2:]:
%o ............if bit-old!=diff:
%o ...............go=False
%o ...............break
%o ............old = bit
%o .........if go:
%o ............return True
%o .
%o for n in count(1):
%o ...if check(n,2):
%o ......print(n)
%o (PARI) isoneap(vch) = {r = vch[2] - vch[1]; for (i=3, #vch, if (vch[i] - vch[i-1] != r, return (0));); return (1);}
%o isap(vd, nch, nd) = {npch = nd/nch; vch = vector(nch); ich = 1; inew = 1; for (i=1, nd, if (inew, vch[ich] = vd[i]; inew = 0;, vch[ich] = 10*vch[ich] + vd[i]); if ((i % npch) == 0, ich++; inew = 1);); isoneap(vch);}
%o isok(n) = {vd = digits(n^2); nd = #vd; if (isprime(nd), return(0)); ok = 0; fordiv(nd, nch, if (nch > 2, if(isap(vd, nch, nd), return (1)));); return (0);} \\ _Michel Marcus_, Jul 06 2014
%K nonn,base
%O 1,1
%A _Christian Perfect_, Jul 04 2014