A202089 Numbers n such that n^2 and (n+1)^2 have same digit sum.

4, 13, 22, 49, 58, 76, 103, 130, 139, 157, 193, 202, 229, 247, 256, 274, 283, 301, 391, 418, 427, 454, 463, 472, 481, 508, 526, 553, 598, 607, 616, 643, 661, 679, 688, 724, 733, 742, 760, 769, 778, 796, 850, 868, 877, 886, 904, 913, 931, 949, 958, 976, 1003

Offset: 1

Author

Zak Seidov, Dec 11 2011

Keywords

• nonn
• base

Comments

Or numbers n such that A004159(n)=A004159(n+1), or A007953(n^2)=A007953((n+1)^2).

Corresponding digit sums are of the form 7+9k, with k=1, 2, 3,... .

Numbers n are of the form 4+9m, with m=0, 1, 2, 5, 6, 8, 11, ... .

A240752(a(n)) = 0. - Reinhard Zumkeller, Apr 12 2014

Examples

			4^2=16 and 5^2=25 have same digit sum ds=7.
13^2=169 and 14^2=196 have ds=16.
76^2=5776 and 77^2=5929 have ds=25.
526^2=276676 and 527^2=277729 have ds=34.
		

Links

• Zak Seidov, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A004159, A007953.

Cf. A239878, A240754.

Programs

Haskell

import Data.List (elemIndices)
a202089 n = a202089_list !! (n-1)
a202089_list = elemIndices 0 a240752_list
-- Reinhard Zumkeller, Apr 12 2014

Mathematica

cnt = 0; nn = 10000; n = 4; Reap[While[cnt < nn, While[Total[IntegerDigits[n^2]] != Total[IntegerDigits[(n + 1)^2]], n = n + 9]; cnt++; Sow[n]; n = n + 9]][[2, 1]]

Python

def ok(n): return sum(map(int, str(n*n))) == sum(map(int, str((n+1)**2)))
print(list(filter(ok, range(1004)))) # Michael S. Branicky, Apr 13 2021