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 A135027 Numbers k such that the sum of the digits of k^2 is 10. Multiples of 10 are omitted. 3
 8, 19, 35, 46, 55, 71, 145, 152, 179, 251, 332, 361, 449, 451, 548, 649, 4499, 20249, 20251, 24499, 100549, 114499, 316261 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS A subsequence of A056020. - R. J. Mathar, Feb 10 2008 Next term > 10000000. - R. J. Mathar, Oct 20 2009 If it exists, a(24) > 10^10. - Hugo Pfoertner, May 17 2021 If it exists, a(24) > 10^29. - Michael S. Branicky, May 30 2021 LINKS Michael S. Branicky, Python program EXAMPLE Corresponding squares are 64, 361, 1225, 2116, 3025, 5041, 21025, 23104, 32041, 63001, 110224, 130321, 201601, 203401, 300304, 421201, 20241001, 410022001, 410103001, 600201001, 10110101401, 13110021001, 100021020121. 8^2 = 64 and 6+4 = 10. 316261^2 = 100021020121 and 1+0+0+0+2+1+0+2+0+1+2+1 = 10. - Zak Seidov, Aug 26 2009 MATHEMATICA s={}; Do[If[Mod[n, 10]>0&&10==Total[IntegerDigits[n^2]], AppendTo[s, n]], {n, 10^8}]; s (* Zak Seidov, Aug 26 2009 *) PROG (Python) def A007953(n): a=0 sh=n while sh > 0: a += sh % 10 sh //= 10 return a def isA135027(n): if n % 10 == 0: return False else: return A007953(n**2) == 10 for n in range(70000): if isA135027(n): print(n) # R. J. Mathar, Oct 20 2009 (Python) # See linked program to go to large numbers def ok(n): return n%10 != 0 and sum(map(int, str(n*n))) == 10 print(list(filter(ok, range(316262)))) # Michael S. Branicky, May 30 2021 (PARI) is(n) = sumdigits(n^2)==10 && n%10 > 0 \\ Felix Fröhlich, May 17 2021 CROSSREFS Cf. A007953, A056020. Sequence in context: A140672 A230098 A262713 * A158916 A045557 A289877 Adjacent sequences: A135024 A135025 A135026 * A135028 A135029 A135030 KEYWORD base,more,nonn AUTHOR Zak Seidov, Feb 10 2008 STATUS approved

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Last modified January 28 21:40 EST 2023. Contains 359905 sequences. (Running on oeis4.)