A356648 -id:A356648

Search: a356648 -id:a356648

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A358880

Squares of the form k + reverse(k) for at least one k.

+10
1

4, 16, 121, 484, 625, 1089, 10201, 14641, 19881, 40804, 49284, 69696, 91809, 94249, 203401, 698896, 1002001, 1234321, 1490841, 1517824, 4008004, 4276624, 4460544, 4937284, 5313025, 6325225, 6895876, 6948496, 7706176, 9018009, 15665764, 15776784, 16120225

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

LINKS

Table of n, a(n) for n=1..33.

FORMULA

a(n) = A356648(n)^2. - Michel Marcus, Dec 25 2022

EXAMPLE

56 + reverse(56) = 56 + 65 = 121 = 11^2, so 121 is a term.

PROG

(Python)

from math import isqrt

def aupto(lim): return sorted(set(t for t in (k + int(str(k)[::-1]) for k in range(1, lim+1)) if t <= lim and isqrt(t)**2 == t))

print(aupto(10**7)) # Michael S. Branicky, Dec 25 2022

CROSSREFS

Cf. A000290, A067030, A356648.

KEYWORD

nonn,base

AUTHOR

Jon E. Schoenfield, Dec 04 2022

STATUS

approved

A358984

The number of n-digit numbers k such that k + digit reversal of k (A056964) is a square.

+10
1

3, 8, 19, 0, 169, 896, 1496, 3334, 21789, 79403, 239439, 651236, 1670022, 3015650, 27292097, 55608749, 234846164, 366081231, 2594727780, 6395506991

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

Number of terms of A061230 which are n digits long.

LINKS

Table of n, a(n) for n=1..20.

Nicolay Avilov, Problem of the Moscow Mathematical Olympiad, 1945 (in Russian).

Index to sequences related to Olympiads.

Index entries for sequences related to Reverse and Add!

EXAMPLE

a(1) = 3 because there are 3 single-digit numbers: 0, 2, 8 such that b + b = m^2, for example, 8 + 8 = 16 = 4^2;

a(2) = 8 because there are 8 two-digit numbers: 29, 38, 47, 56, 65, 74, 83, 92 such that bc + cb = m^2, for example, 29 + 92 = 121 = 11^2.

MATHEMATICA

a[n_]:=Length[Select[Table[k, {k, 10^(n-1), 10^n-1}], IntegerQ[Sqrt[#+FromDigits[Reverse[IntegerDigits[#]]]]]&]]; Array[a, 10] (* Stefano Spezia, Dec 09 2022 *)

PROG

(Python)

from math import isqrt

def s(n): return isqrt(n)**2 == n

def c(n): return s(n + int(str(n)[::-1]))

def a(n): return 3 if n == 1 else sum(1 for k in range(10**(n-1), 10**n) if c(k))

print([a(n) for n in range(1, 7)]) # Michael S. Branicky, Dec 08 2022

CROSSREFS

Cf. A056964, A061230, A356648.

KEYWORD

nonn,base,more

AUTHOR

Nicolay Avilov, Dec 08 2022

EXTENSIONS

a(9)-a(10) from Michael S. Branicky, Dec 08 2022

a(11)-a(20) from Talmon Silver, Dec 25 2022

STATUS

approved

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