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A135027
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Numbers k such that the sum of the digits of k^2 is 10. Multiples of 10 are omitted.
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3
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8, 19, 35, 46, 55, 71, 145, 152, 179, 251, 332, 361, 449, 451, 548, 649, 4499, 20249, 20251, 24499, 100549, 114499, 316261
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listen;
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internal format)
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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Table of n, a(n) for n=1..23.
Michael S. Branicky, Python program
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EXAMPLE
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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
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MATHEMATICA
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s={}; Do[If[Mod[n, 10]>0&&10==Total[IntegerDigits[n^2]], AppendTo[s, n]], {n, 10^8}]; s (* Zak Seidov, Aug 26 2009 *)
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PROG
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(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
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CROSSREFS
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Cf. A007953, A056020.
Sequence in context: A140672 A230098 A262713 * A158916 A045557 A289877
Adjacent sequences: A135024 A135025 A135026 * A135028 A135029 A135030
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KEYWORD
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base,more,nonn
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AUTHOR
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Zak Seidov, Feb 10 2008
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STATUS
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approved
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