cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A214492 Arithmetic mean of next a(n) successive squares of positive integers is a square.

Original entry on oeis.org

337, 649, 961, 1273, 1585, 1897, 1919, 1151, 1223, 1295, 1367, 1439, 1511, 1583, 1655, 1727, 1799, 1871, 1943, 2015, 2087, 2159, 2231, 2303, 2375, 2447, 2519, 1487, 1511, 1535, 1559, 1583, 1607, 1631, 1655, 1679, 1703, 1727, 1751, 1775, 1799, 1823, 1847, 1871, 1895, 1919
Offset: 1

Views

Author

Alex Ratushnyak, Jul 19 2012

Keywords

Comments

Only squares of positive integers, starting from 1; zero is not included. (If it were included, A103214 would result.)
Also, a(n)=1 is obviously not permitted: must be 2 or more successive squares, otherwise all a(n)=1.
Among first 1363 terms all are odd, 933 are primes, a(n) < a(n-1) twice.
Corresponding arithmetic means that are perfect squares:
b(n) = 38025, 473344, 2229049, 6812100, 16313521, 33408400, 59013124, 84695209, 107952100, 135699201, 168480400, 206870689, 251476164, 302934025, 361912576, 429111225, 505260484, 591121969, 687488400
Their square roots c(n) = sqrt(b(n)):
195, 688, 1493, 2610, 4039, 5780, 7682, 9203, 10390, 11649, 12980, 14383, 15858, 17405, 19024, 20715, 22478, 24313, 26220, 28199, 30250, 32373, 34568, 36835, 39174, 41585, 44068, 46067, 47566, 49089, 50636, 52207

Examples

			(1 + 4 + 9 + ... + 337^2)/337 = 38025, which is a square, so 337 is a term.
(338^2 + ... + (338 + 648)^2)/649 = 473344, which is a square, so 649 is a term.
		

Crossrefs

Cf. A103214, excluding first term: arithmetic mean of next a(n) successive squares of nonnegative integers is a square.

Programs

  • Python
    import math
    sum = k = 0
    for n in range(1, 220000):
        sum += n*n
        k += 1
        sqr = int(math.sqrt(sum*1.0/k))
        while sqr*sqr*ksum:
            sqr-=1
        if sqr*sqr*k==sum and k>1:
            print(k, end=',')
            sum = k = 0