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.

Previous Showing 31-37 of 37 results.

A004814 Numbers that are the sum of 3 positive 11th powers.

Original entry on oeis.org

3, 2050, 4097, 6144, 177149, 179196, 181243, 354295, 356342, 531441, 4194306, 4196353, 4198400, 4371452, 4373499, 4548598, 8388609, 8390656, 8565755, 12582912, 48828127, 48830174, 48832221, 49005273, 49007320, 49182419, 53022430
Offset: 1

Views

Author

Keywords

Comments

As the order of addition doesn't matter we can assume terms are in nondecreasing order. - David A. Corneth, Aug 01 2020

Examples

			From _David A. Corneth_, Aug 01 2020: (Start)
204800049005272 is in the sequence as 204800049005272 = 3^11 + 5^11 + 20^11.
2518268235958260 is in the sequence as 2518268235958260 = 16^11 + 19^11 + 25^11.
3786934745885995 is in the sequence as 3786934745885995 = 10^11 + 19^11 + 26^11. (End)
		

Crossrefs

A###### (x, y): Numbers that are the form of x nonzero y-th powers.
Cf. A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).

Extensions

Removed incorrect program. - David A. Corneth, Aug 01 2020

A186665 Total number of positive integers below 10^n requiring 10 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 6, 72, 812, 7876, 70630, 673358, 6676098, 66678723, 666681488, 6666684311
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A114322(n) + A186649(n) + A186651(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + a(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n).

Crossrefs

Extensions

a(5)-a(7) from Lars Blomberg, May 08 2011
a(8)-a(9) from Hiroaki Yamanouchi, Oct 13 2014
a(10)-a(11) from Giovanni Resta, Apr 26 2016

A186666 Total number of n-digit numbers requiring 10 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 6, 66, 740, 7064, 62754, 602728, 6002740, 60002625, 600002765, 6000002823
Offset: 1

Views

Author

Martin Renner, Feb 25 2011

Keywords

Comments

A102831(n) + A186650(n) + A186652(n) + A186654(n) + A186656(n) + A186658(n) + A186660(n) + A186662(n) + A186664(n) + a(n) + A186668(n) + A186670(n) + A186672(n) + A186674(n) + A186676(n) + A186678(n) + A186681(n) + A186683(n) + A186685(n) = A052268(n), for n>1.

Crossrefs

Formula

a(n) = A186665(n) - A186665(n-1).

Extensions

a(5)-a(11) from Giovanni Resta, Apr 29 2016

A345595 Numbers that are the sum of ten fourth powers in two or more ways.

Original entry on oeis.org

265, 280, 295, 310, 325, 340, 345, 355, 360, 375, 390, 405, 420, 425, 440, 455, 470, 485, 505, 520, 535, 550, 565, 580, 585, 595, 600, 615, 630, 645, 660, 665, 680, 695, 710, 725, 745, 760, 775, 790, 805, 820, 835, 840, 855, 870, 885, 889, 900, 904, 919, 920
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			280 is a term because 280 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**4 for x in range(1, 1000)]
    for pos in cwr(power_terms, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 2])
        for x in range(len(rets)):
            print(rets[x])

A345853 Numbers that are the sum of ten fourth powers in exactly one ways.

Original entry on oeis.org

10, 25, 40, 55, 70, 85, 90, 100, 105, 115, 120, 130, 135, 145, 150, 160, 165, 170, 180, 185, 195, 200, 210, 215, 225, 230, 245, 250, 260, 275, 290, 330, 370, 385, 400, 410, 435, 450, 465, 490, 500, 515, 530, 570, 610, 625, 634, 640, 649, 650, 664, 675, 679
Offset: 1

Views

Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A003344 at term 30 because 265 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4.

Examples

			25 is a term because 25 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**4 for x in range(1, 1000)]
    for pos in cwr(power_terms, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 1])
        for x in range(len(rets)):
            print(rets[x])

A003386 Numbers that are the sum of 8 nonzero 8th powers.

Original entry on oeis.org

8, 263, 518, 773, 1028, 1283, 1538, 1793, 2048, 6568, 6823, 7078, 7333, 7588, 7843, 8098, 8353, 13128, 13383, 13638, 13893, 14148, 14403, 14658, 19688, 19943, 20198, 20453, 20708, 20963, 26248, 26503, 26758, 27013, 27268, 32808, 33063, 33318, 33573
Offset: 1

Views

Author

Keywords

Comments

As the order of addition doesn't matter we can assume terms are in nondecreasing order. - David A. Corneth, Aug 01 2020

Examples

			From _David A. Corneth_, Aug 01 2020: (Start)
9534597 is in the sequence as 9534597 = 2^8 + 3^8 + 3^8 + 3^8 + 5^8 + 6^8 + 6^8 + 7^8.
13209988 is in the sequence as 13209988 = 1^8 + 1^8 + 2^8 + 2^8 + 2^8 + 6^8 + 7^8 + 7^8.
19046628 is in the sequence as 19046628 = 2^8 + 2^8 + 3^8 + 4^8 + 6^8 + 7^8 + 7^8 + 7^8. (End)
		

Crossrefs

A###### (x, y): Numbers that are the form of x nonzero y-th powers.
Cf. A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).

Programs

  • Mathematica
    M = 92646056; m = M^(1/8) // Ceiling; Reap[
    For[a = 1, a <= m, a++, For[b = a, b <= m, b++, For[c = b, c <= m, c++,
    For[d = c, d <= m, d++, For[e = d, e <= m, e++, For[f = e, f <= m, f++,
    For[g = f, g <= m, g++, For[h = g, h <= m, h++,
    s = a^8 + b^8 + c^8 + d^8 + e^8 + f^8 + g^8 + h^8;
    If[s <= M, Sow[s]]]]]]]]]]][[2, 1]] // Union (* Jean-François Alcover, Dec 01 2020 *)

Extensions

b-file checked by R. J. Mathar, Aug 01 2020
Incorrect program removed by David A. Corneth, Aug 01 2020

A047721 Sum of 10 but no fewer nonzero fourth powers.

Original entry on oeis.org

10, 25, 40, 55, 70, 90, 105, 120, 135, 150, 160, 170, 185, 200, 215, 225, 230, 250, 265, 280, 295, 310, 330, 345, 360, 375, 390, 400, 410, 425, 440, 455, 465, 470, 485, 490, 505, 520, 535, 550, 565, 570, 585, 600, 615, 634, 640, 650, 665, 680, 695, 714, 730
Offset: 1

Views

Author

Arlin Anderson (starship1(AT)gmail.com)

Keywords

Crossrefs

Programs

  • PARI
    upto(n)={my(e=10); my(s=sum(k=1, sqrtint(sqrtint(n)), x^(k^4)) + O(x*x^n)); my(p=s^e, q=(1 + s)^(e-1)); select(k->polcoeff(p,k) && !polcoeff(q,k), [1..n])} \\ Andrew Howroyd, Jul 06 2018
    
  • Python
    from itertools import count, takewhile, combinations_with_replacement as mc
    def aupto(lim):
        p4 = list(takewhile(lambda x: x <= lim, (i**4 for i in count(1))))
        s = [set(sum(c) for c in mc(p4, i) if sum(c) <= lim) for i in range(11)]
        ans = s[10]
        for i in range(1, 10):
            ans -= s[i]
        return sorted(ans)
    print(aupto(730)) # Michael S. Branicky, Oct 25 2021
Previous Showing 31-37 of 37 results.