A002283 -id:A002283 - cp's OEIS frontend

A130130 a(0)=0, a(1)=1, a(n)=2 for n >= 2.

0, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

Paul Curtz, Aug 01 2007

a(n) is also total number of positive integers below 10^(n+1) requiring 9 positive cubes in their representation as sum of cubes (cf. Dickson, 1939).
A061439(n) + A181375(n) + A181377(n) + A181379(n) + A181381(n) + A181400(n) + A181402(n) + A181404(n) + a(n) = A002283(n).
a(n) = number of obvious divisors of n. The obvious divisors of n are the numbers 1 and n. - Jaroslav Krizek, Mar 02 2009
Number of colors needed to paint n adjacent segments on a line. - Jaume Oliver Lafont, Mar 20 2009
a(n) = ceiling(n-th nonprimes/n) = ceiling(A018252(n)/A000027(n)) for n >= 1. Numerators of (A018252(n)/A000027(n)) in A171529(n), denominators of (A018252(n)/A000027(n)) in A171530(n). a(n) = A171624(n) + 1 for n >= 5. - Jaroslav Krizek, Dec 13 2009
a(n) is also the continued fraction for sqrt(1/2). - Enrique Pérez Herrero, Jul 12 2010
For n >= 1, a(n) = minimal number of divisors of any n-digit number. See A066150 for maximal number of divisors of any n-digit number. - Jaroslav Krizek, Jul 18 2010
Central terms in the triangle A051010. - Reinhard Zumkeller, Jun 27 2013
Decimal expansion of 11/900. - Elmo R. Oliveira, May 05 2024

Leonard Eugene Dickson, All integers except 23 and 239 are sums of eight cubes, Bulletin of the American Mathematical Society 45 (1939), p. 588-591.
Eric Weisstein's World of Mathematics, Waring's Problem.
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A000005, A000027, A002283, A018252, A051010, A061439, A066150, A070824, A158411, A171529, A171530, A171624, A181375, A181377, A181379, A181381, A181400, A181402, A181404.

a130130 = min 2
a130130_list = 0 : 1 : repeat 2  -- Reinhard Zumkeller, Jun 27 2013

A130130[0]:=0; A130130[1]:=1; A130130[n_]:=2; (* Enrique Pérez Herrero, Jul 12 2010 *)
A130130[n_]:=ContinuedFraction[Sqrt[1/2],n+1][[n+1]] (* Enrique Pérez Herrero, Jul 12 2010 *)
Join[{0, 1},LinearRecurrence[{1},{2},96]] (* Ray Chandler, Sep 23 2015 *)
PadRight[{0,1},120,{2}] (* Harvey P. Dale, Sep 15 2022 *)

a(n)=min(n,2) \\ Charles R Greathouse IV, Jun 01 2011

G.f.: x*(1+x)/(1-x) = x*(1-x^2)/(1-x)^2. - Jaume Oliver Lafont, Mar 20 2009
a(n) = A000005(n) - A070824(n) for n >= 1.
E.g.f.: 2*exp(x) - x - 2. - Stefano Spezia, May 19 2024

A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

Original entry on oeis.org

5, 10, 13, 17, 20, 25, 26, 29, 34, 41, 37, 40, 45, 52, 61, 50, 53, 58, 65, 74, 85, 65, 68, 73, 80, 89, 100, 113, 82, 85, 90, 97, 106, 117, 130, 145, 101, 104, 109, 116, 125, 136, 149, 164, 181, 122, 125, 130, 137, 146, 157, 170, 185, 202, 221, 145, 148, 153, 160

Offset: 2

Antti Karttunen, Apr 04 2000

Discovered by Bernard Frénicle de Bessy (1605?-1675). - Paul Curtz, Aug 18 2008
Terms that are not hypotenuses in primitive Pythagorean triangles, are replaced by 0 in A222946. - Reinhard Zumkeller, Mar 23 2013
This triangle T(n,k) gives the circumdiameters for the Pythagorean triangles with a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (see the Floor van Lamoen entries or comments A063929, A063930, A002283, A003991). See also the formula section. Note that not all Pythagorean triangles are covered, e.g., (9,12,15) does not appear. - Wolfdieter Lang, Dec 03 2014

			The triangle T(n, k) begins:
n\k   1   2   3   4   5   6   7   8   9  10  11 ...
2:    5
3:   10  13
4:   17  20  25
5:   26  29  34  41
6:   37  40  45  52  61
7:   50  53  58  65  74  85
8:   65  68  73  80  89 100 113
9:   82  85  90  97 106 117 130 145
10: 101 104 109 116 125 136 149 164 181
11: 122 125 130 137 146 157 170 185 202 221
12: 145 148 153 160 169 180 193 208 225 244 265
...
13: 170 173 178 185 194 205 218 233 250 269 290 313,
14: 197 200 205 212 221 232 245 260 277 296 317 340 365,
15: 226 229 234 241 250 261 274 289 306 325 346 369 394 421,
16: 257 260 265 272 281 292 305 320 337 356 377 400 425 452 481,
...
Formatted and extended by _Wolfdieter Lang_, Dec 02 2014 (reformatted Jun 11 2015)
The successive terms are (1^2+2^2), (1^2+3^2), (2^2+3^2), (1^2+4^2), (2^2+4^2), (3^2+4^2), ...

Reinhard Zumkeller, Rows n = 2..121 of triangle, flattened
M. de Frénicle, Méthode pour trouver la solutions des problèmes par les exclusions, in: "Divers ouvrages de mathématiques et de physique, par Messieurs de l'Académie royale des sciences", Paris, 1693, pp 1-44.
Antti Karttunen, Larger table, showing also locations of 4k+1 primes and squares
Eric Weisstein's World of Mathematics, Congruum Problem.
Index entries for sequences related to sums of squares

Sorting gives A024507. Count of divisors: A055097, Möbius: A055132. For trinv, follow A055088.
Cf. A001844 (right edge), A002522 (left edge), A033429 (central column).
Cf. A000290, A033583, A079273, A190816, A226141, A331987.

a055096 n k = a055096_tabl !! (n-1) !! (k-1)
a055096_row n = a055096_tabl !! (n-1)
a055096_tabl = zipWith (zipWith (+)) a133819_tabl a140978_tabl
-- Reinhard Zumkeller, Mar 23 2013

[n^2+k^2: k in [1..n-1], n in [2..15]]; // G. C. Greubel, Apr 19 2023

sum2distinct_squares_array := (n) -> (((n-((trinv(n-1)*(trinv(n-1)-1))/2))^2)+((trinv(n-1)+1)^2));

T[n_, k_]:= (n+1)^2 + k^2; Table[T[n, k], {n,15}, {k,n}]//Flatten (* Jean-François Alcover, Mar 16 2015, after Reinhard Zumkeller *)

def A055096(n,k): return n^2 + k^2
flatten([[A055096(n,k) for k in range(1,n)] for n in range(2,16)]) # G. C. Greubel, Apr 19 2023

a(n) = sum2distinct_squares_array(n).
T(n, 1) = A002522(n).
T(n, n-1) = A001844(n-1).
T(2*n-2, n-1) = A033429(n-1).
T(n,k) = A133819(n,k) + A140978(n,k) = (n+1)^2 + k^2, 1 <= k <= n. - Reinhard Zumkeller, Mar 23 2013
T(n, k) = a*b*c/(2*sqrt(s*(s-1)*(s-b)*(s-c))) with s =(a + b + c)/2 and the substitution a = (n+1)^2 - k^2, b = 2*(n+1)*k and c = (n+1)^2 + k^2 (the circumdiameter for the considered Pythagorean triangles). - Wolfdieter Lang, Dec 03 2014
From Bob Selcoe, Mar 21 2015:  (Start)
T(n,k) = 1 + (n-k+1)^2 + Sum_{j=0..k-2} (4*j + 2*(n-k+3)).
T(n,k) = 1 + (n+k-1)^2 - Sum_{j=0..k-2} (2*(n+k-3) - 4*j).
Therefore: 4*(n-k+1) + Sum_{j=0..k-2} (2*(n-k+3) + 4*j) = 4*n(k-1) - Sum_{j=0..k-2} (2*(n+k-3) - 4*j). (End)
From G. C. Greubel, Apr 19 2023: (Start)
T(2*n-3, n-1) = A033429(n-1).
T(2*n-4, n-2) = A079273(n-1).
T(2*n-2, n) = A190816(n).
T(3*n-4, n-1) = 10*A000290(n-1) = A033583(n-1).
Sum_{k=1..n-1} T(n, k) = A331987(n-1).
Sum_{k=1..floor(n/2)} T(n-k, k) = A226141(n-1). (End)

Edited: in T(n, k) formula by Reinhard Zumkeller k < n replaced by k <= n. - Wolfdieter Lang, Dec 02 2014

Made definition more precise, changed offset to 2. - N. J. A. Sloane, Mar 30 2015

A058369 Numbers k such that k and k^2 have same digit sum.

Original entry on oeis.org

0, 1, 9, 10, 18, 19, 45, 46, 55, 90, 99, 100, 145, 180, 189, 190, 198, 199, 289, 351, 361, 369, 379, 388, 450, 451, 459, 460, 468, 495, 496, 550, 558, 559, 568, 585, 595, 639, 729, 739, 775, 838, 855, 900, 954, 955, 990, 999, 1000, 1098, 1099, 1179, 1188, 1189

Offset: 1

G. L. Honaker, Jr., Dec 17 2000

It is interesting that the graph of this sequence appears almost identical as the maximum value of n increases by factors of 10. Compare the graph of the b-file (having numbers up to 10^6) with the plot of the terms up to 10^8. - T. D. Noe, Apr 28 2012
If iterated digit sum (A010888, A056992) is used instead of just digit sum (A007953, A004159), we get A090570 of which this sequence is a subset. - Jeppe Stig Nielsen, Feb 18 2015
Hare, Laishram, & Stoll show that this sequence (indeed, even its subsequence A254066) is infinite. In particular for each k in {846, 847, 855, 856, 864, 865, 873, ...} there are infinitely many terms in this sequence not divisible by 10 that have digit sum k. - Charles R Greathouse IV, Aug 25 2015
There are infinitely many n such that both n and n+1 are in the sequence.  This includes A002283. - Robert Israel, Aug 26 2015

			Digit sum of 9 = 9 9^2 = 81, 8+1 = 9 digit sum of 145 = 1+4+5 = 10 145^2 = 21025, 2+1+0+2+5 = 10 digit sum of 954 = 9+5+4 = 18 954^2 = 910116, 9+1+0+1+1+6 = 18. - Florian Roeseler (hazz_dollazz(AT)web.de), May 03 2010

Zak Seidov, Table of n, a(n) for n = 1..8354 (to a(n) = 10^6)
Code Golf StackExchange, Equality in the sum of digits, coding challenge started Mar 11 2016.
K. G. Hare, S. Laishram, and T. Stoll, The sum of digits of n and n^2, International Journal of Number Theory 7:7 (2011), pp. 1737-1752.
T. D. Noe, Plot of terms up to 10^8

Cf. A002283, A007953, A058370, A058852, A077436, A254066.
Cf. A147523 (number of numbers in each decade).
Subsequence of A090570.

import Data.List (elemIndices)
import Data.Function (on)
a058369 n = a058369_list !! (n-1)
a058369_list =
   elemIndices 0 $ zipWith ((-) `on` a007953) [0..] a000290_list
-- Reinhard Zumkeller, Aug 17 2011

[n: n in [0..1200] |(&+Intseq(n)) eq (&+Intseq(n^2))]; // Vincenzo Librandi, Aug 26 2015

sd := proc (n) options operator, arrow: add(convert(n, base, 10)[j], j = 1 .. nops(convert(n, base, 10))) end proc: a := proc (n) if sd(n) = sd(n^2) then n else end if end proc; seq(a(n), n = 0 .. 1400); # Emeric Deutsch, May 11 2010
select(t -> convert(convert(t,base,10),`+`)=convert(convert(t^2,base,10),`+`),
[seq(seq(9*i+j,j=0..1),i=0..1000)]); # Robert Israel, Aug 26 2015

Select[Range[0,1200],Total[IntegerDigits[#]]==Total[IntegerDigits[ #^2]]&] (* Harvey P. Dale, Jun 14 2011 *)

is(n)=sumdigits(n)==sumdigits(n^2) \\ Charles R Greathouse IV, Aug 25 2015

def ds(n): return sum(map(int, str(n)))
def ok(n): return ds(n) == ds(n**2)
def aupto(nn): return [m for m in range(nn+1) if ok(m)]
print(aupto(1189)) # Michael S. Branicky, Jan 09 2021

A007953(a(n)) = A004159(a(n)). - Reinhard Zumkeller, Apr 25 2009

Edited by N. J. A. Sloane, May 30 2010

A059988 a(n) = (10^n - 1)^2.

Original entry on oeis.org

0, 81, 9801, 998001, 99980001, 9999800001, 999998000001, 99999980000001, 9999999800000001, 999999998000000001, 99999999980000000001, 9999999999800000000001, 999999999998000000000001, 99999999999980000000000001, 9999999999999800000000000001, 999999999999998000000000000001

Offset: 0

Henry Bottomley, Mar 07 2001

From James D. Klein, Feb 05 2012: (Start)
The periods of the reciprocals of a(n) are the consecutive integers from 0 to 10^n-1, omitting the one integer 10^n-2, right-justified in field widths of size n.
E.g.:
  1/81 = 0.012345679...
  1/9801 = 0.000102030405060708091011...9799000102...
  1/998001 = 0.000001002003004005...997999000001002... (End)
Sum of first 10^n - 1 odd numbers. - Arkadiusz Wesolowski, Jun 12 2013

			From _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ..................... 81 = 9^2;
n=2: ................... 9801 = 99^2;
n=3: ................. 998001 = 999^2;
n=4: ............... 99980001 = 9999^2;
n=5: ............. 9999800001 = 99999^2;
n=6: ........... 999998000001 = 999999^2;
n=7: ......... 99999980000001 = 9999999^2;
n=8: ....... 9999999800000001 = 99999999^2;
n=9: ..... 999999998000000001 = 999999999^2. (End)

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 32 at p. 61.
Walther Lietzmann, Lustiges und Merkwuerdiges von Zahlen und Formen, (F. Hirt, Breslau 1921-43), p. 149.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 34.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A178630, A178631, A178632, A178633, A178634, A178635, A272066, A272067, A272068.
Subsequence of A238237.
Cf. A002275, A002283, A002477.

A059988:=n->(10^n-1)^2; seq(A059988(n), n=0..20); # Wesley Ivan Hurt, Nov 21 2013

Table[(10^n-1)^2, {n,0,20}] (* Wesley Ivan Hurt, Nov 21 2013 *)

a(n) = { (10^n - 1)^2 } \\ Harry J. Smith, Jul 01 2009

def a(n): return (10**n - 1)**2  # Martin Gergov, Sep 10 2022

a(n) = 81*A002477(n) = A002283(n)^2 = (9*A002275(n))^2.
a(n) = {999... (n times)}^2 = {999... (n times), 000... (n times)} - {999... (n times)}. For example, 999^2 = 999000 - 999 = 998001. - Kyle D. Balliet, Mar 07 2009
a(n) = (A002283(n-1)*10 + 8) * 10^(n-1) + 1, for n>0. - Reinhard Zumkeller, May 31 2010
From Ilya Gutkovskiy, Apr 19 2016: (Start)
O.g.f.: 81*x*(1 + 10*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 2*exp(9*x) + exp(99*x))*exp(x). (End)
Sum_{n>=1} 1/a(n) = (log(10)*(QPolyGamma(0, 1, 1/10) - log(10/9)) + QPolyGamma(1, 1, 1/10))/log(10)^2 = 0.012448721523422795191... . - Stefano Spezia, Jul 31 2024
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). - Elmo R. Oliveira, Aug 02 2025

A075412 Squares of A002277.

Original entry on oeis.org

0, 9, 1089, 110889, 11108889, 1111088889, 111110888889, 11111108888889, 1111111088888889, 111111110888888889, 11111111108888888889, 1111111111088888888889, 111111111110888888888889, 11111111111108888888888889, 1111111111111088888888888889, 111111111111110888888888888889

Offset: 0

Michael Taylor (michael.taylor(AT)vf.vodafone.co.uk), Sep 14 2002

A transformation of the Wonderful Demlo numbers (A002477).

			a(2) = 33^2 = 1089.
Contribution from _Reinhard Zumkeller_, May 31 2010: (Start)
n=1: ...................... 9 = 9 * 1;
n=2: ................... 1089 = 99 * 11;
n=3: ................. 110889 = 999 * 111;
n=4: ............... 11108889 = 9999 * 1111;
n=5: ............. 1111088889 = 99999 * 11111;
n=6: ........... 111110888889 = 999999 * 111111;
n=7: ......... 11111108888889 = 9999999 * 1111111;
n=8: ....... 1111111088888889 = 99999999 * 11111111;
n=9: ..... 111111110888888889 = 999999999 * 111111111. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Gérard Villemin, Variations sur les carrés.
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A075411, A075412, A075413, A075414, A075415, A075416, A075417, A002283, A178630, A178631, A178632, A178633, A178634, A178635.

Cf. A000042, A002277, A002275, A002282, A002283, A002477, A059988.

LinearRecurrence[{11, -10}, {0, 3}, 20]^2 (* Vincenzo Librandi, Mar 20 2014 *)
Table[FromDigits[PadRight[{},n,9]]FromDigits[PadRight[{},n,1]],{n,0,15}] (* Harvey P. Dale, Feb 12 2023 *)

a(n) = A002277(n)^2 = (3*A002275(n))^2 = 9*A002275(n)^2.
a(n) = {111111... (2n times)} - 2*{ 111... (n times)} a(n) = A000042(2*n) - 2*A000042(n). - Amarnath Murthy, Jul 21 2003
a(n) = {333... (n times)}^2 = {111...(n times)}{000... (n times)} - {111... (n times)}. For example, 333^2 = 111000 - 111 = 110889. - Kyle D. Balliet, Mar 07 2009
From Reinhard Zumkeller, May 31 2010: (Start)
a(n) = A002283(n)*A002275(n).
For n>0, a(n) = (A002275(n-1)*10^n + A002282(n-1))*10 + 9. (End)
a(n) = (10^(n+1)-10)^2/900. - José de Jesús Camacho Medina, Apr 01 2016
From Elmo R. Oliveira, Jul 27 2025: (Start)
G.f.: 9*x*(1+10*x)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 - 2*exp(9*x) + exp(99*x))/9.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3).
a(n) = 9*A002477(n). (End)

A186684 Total number of positive integers below 10^n requiring 19 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

0, 1, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 1

Martin Renner, Feb 25 2011

A114322(n) + A186649(n) + A186651(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + a(n) = A002283(n).

J.-M. Deshouillers, K. Kawada, and T. D. Wooley, On sums of sixteen biquadrates, Mem. Soc. Math. Fr. 100 (2005), p. 120.

J.-M. Deshouillers, F. Hennecart and B. Landreau, Waring's Problem for sixteen biquadrates - numerical results, Journal de Théorie des Nombres de Bordeaux 12 (2000), pp. 411-422.
L. E. Dickson, Recent progress on Waring's theorem and its generalizations, Bull. Amer. Math. Soc. 39:10 (1933), pp. 701-727.
Eric Weisstein's World of Mathematics, Waring's Problem.
Index entries for linear recurrences with constant coefficients, signature (1).

Cf. A010727, A046050.

PadRight[{0, 1}, 100, 7] (* Paolo Xausa, Jul 30 2024 *)

a(n) = 7 for n >= 3. - Nathaniel Johnston, May 09 2011
From Elmo R. Oliveira, Aug 05 2024: (Start)
G.f.: x^2*(1 + 6*x)/(1 - x).
E.g.f.: 7*(exp(x) - 1 - x) - 3*x^2. (End)

a(5)-a(6) from Lars Blomberg, May 08 2011

Terms after a(6) from Nathaniel Johnston, May 09 2011

A114322 Largest number whose 4th power has n digits.

Original entry on oeis.org

1, 3, 5, 9, 17, 31, 56, 99, 177, 316, 562, 999, 1778, 3162, 5623, 9999, 17782, 31622, 56234, 99999, 177827, 316227, 562341, 999999, 1778279, 3162277, 5623413, 9999999, 17782794, 31622776, 56234132, 99999999, 177827941, 316227766, 562341325, 999999999, 1778279410

Offset: 1

Jonathan Vos Post, Feb 06 2006

This is to 4th powers as A061439 is to cubes and A049416 is to squares.

a(n) + A186649(n) + A186651(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n).

			a(10) = 316 because 316^4 = 9971220736 which has 10 digits, while 317^4 = 10098039121 has 11 digits.
a(35) = 562341325 because 562341325^4 = 99999999864602459914272843469140625 has 35 digits, while 562341326^4 = 100000000575914225104884587789852176 has 36.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A061439, A049416.

[Ceiling((10^n)^(1/4))-1: n in [1..40]]; // Vincenzo Librandi, Oct 01 2011

Ceiling[(10^Range[50])^(1/4)] - 1 (* Paolo Xausa, Jul 30 2024 *)

a(n) = ceiling((10^n)^(1/4)) - 1.

A186649 Total number of positive integers below 10^n requiring 2 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

1, 5, 14, 43, 143, 460, 1467, 4613, 14629, 46341, 146545, 463344, 1465658, 4634967, 14657277, 46350371

Offset: 1

Martin Renner, Feb 25 2011

nonn

A114322(n) + a(n) + A186651(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n).

Eric Weisstein's World of Mathematics, Waring's Problem.

Cf. A003336.

isbiquadrate:=proc(n) type(root(n,4),posint); end:
isA003336:=proc(n) local x,y4; if isbiquadrate(n) then false; else for x from 1 do y4:=n-x^4; if y4A003336(k) then i:=i+1; fi; od: return(i); end: for n from 1 do print(a(n)); od;

a(6) from Martin Renner, Feb 26 2011

a(7)-a(16) from Lars Blomberg, May 08 2011

A186651 Total number of positive integers below 10^n requiring 3 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

1, 6, 29, 141, 789, 4353, 24173, 134679, 756947, 4258210, 23958106

Offset: 1

Martin Renner, Feb 25 2011

A114322(n) + A186649(n) + a(n) + A186653(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n).

Eric Weisstein's World of Mathematics, Waring's Problem.

Cf. A003337.

Cf. A002283, A114322, A186649, A186653, A186655, A186657, A186659, A186661, A186663, A186665, A186667, A186669, A186671, A186673, A186675, A186677, A186680, A186682, A186684.

a(5)-a(11) from Lars Blomberg, May 08 2011

A186653 Total number of positive integers below 10^n requiring 4 positive biquadrates in their representation as sum of biquadrates.

Original entry on oeis.org

1, 7, 48, 346, 3066, 27754, 260724, 2516312, 24744689, 245221669, 2442288495

Offset: 1

Martin Renner, Feb 25 2011

nonn

A114322(n) + A186649(n) + A186651(n) + a(n) + A186655(n) + A186657(n) + A186659(n) + A186661(n) + A186663(n) + A186665(n) + A186667(n) + A186669(n) + A186671(n) + A186673(n) + A186675(n) + A186677(n) + A186680(n) + A186682(n) + A186684(n) = A002283(n)

Eric Weisstein's World of Mathematics, Waring's Problem.

Cf. A003338, A186654.

a(5)-a(9) from Lars Blomberg, May 08 2011

a(10)-a(11) from Giovanni Resta, Apr 26 2016

cp's OEIS Frontend

A130130 a(0)=0, a(1)=1, a(n)=2 for n >= 2.

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A055096 Triangle read by rows, sums of 2 distinct nonzero squares: T(n,k) = k^2+n^2, (n>=2, 1 <= k <= n-1).

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A058369 Numbers k such that k and k^2 have same digit sum.

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A059988 a(n) = (10^n - 1)^2.

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A075412 Squares of A002277.

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A186684 Total number of positive integers below 10^n requiring 19 positive biquadrates in their representation as sum of biquadrates.

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