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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
Sequence in context: A140672 A230098 A262713 * A158916 A045557 A289877
KEYWORD
base,more,nonn
AUTHOR
Zak Seidov, Feb 10 2008
STATUS
approved

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Last modified March 28 20:05 EDT 2024. Contains 371254 sequences. (Running on oeis4.)