A057945 -id:A057945 - cp's OEIS frontend

A053610 Number of positive squares needed to sum to n using the greedy algorithm.

1, 2, 3, 1, 2, 3, 4, 2, 1, 2, 3, 4, 2, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 5, 2, 1, 2, 3, 4, 2, 3, 4, 5, 3, 2, 3, 4, 5, 3, 4, 5, 2, 3, 4, 1, 2, 3, 4

Offset: 1

Jud McCranie, Mar 19 2000

nonn

Define f(n) = n - x^2 where (x+1)^2 > n >= x^2. a(n) = number of iterations in f(...f(f(n))...) to reach 0.
a(n) = 1 iff n is a perfect square.
Also sum of digits when writing n in base where place values are squares, cf. A007961. - Reinhard Zumkeller, May 08 2011
The sequence could have started with a(0)=0. - Thomas Ordowski, Jul 12 2014
The sequence is not bounded, see A006892. - Thomas Ordowski, Jul 13 2014

			7=4+1+1+1, so 7 requires 4 squares using the greedy algorithm, so a(7)=4.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A006892 (positions of records), A055401, A007961.

Cf. A000196, A000290, A057945 (summing triangular numbers).

a053610 n = s n $ reverse $ takeWhile (<= n) $ tail a000290_list where
  s _ []                 = 0
  s m (x:xs) | x > m     = s m xs
             | otherwise = m' + s r xs where (m',r) = divMod m x
-- Reinhard Zumkeller, May 08 2011

A053610 := proc(n)
    local a,x;
    a := 0 ;
    x := n ;
    while x > 0 do
        x := x-A048760(x) ;
        a := a+1 ;
    end do:
    a ;
end proc: # R. J. Mathar, May 13 2016

f[n_] := (n - Floor[Sqrt[n]]^2); g[n_] := (m = n; c = 1; While[a = f[m]; a != 0, c++; m = a]; c); Table[ g[n], {n, 1, 105}]

A053610(n,c=1)=while(n-=sqrtint(n)^2,c++);c \\ M. F. Hasler, Dec 04 2008

from math import isqrt
def A053610(n):
    c = 0
    while n:
        n -= isqrt(n)**2
        c += 1
    return c # Chai Wah Wu, Aug 01 2023

a(n) = A007953(A007961(n)). - Henry Bottomley, Jun 01 2000
a(n) = a(n - floor(sqrt(n))^2) + 1 = a(A053186(n)) + 1 [with a(0) = 0]. - Henry Bottomley, May 16 2000
A053610 = A002828 + A062535. - M. F. Hasler, Dec 04 2008

A057944 Largest triangular number less than or equal to n; write m-th triangular number m+1 times.

Original entry on oeis.org

0, 1, 1, 3, 3, 3, 6, 6, 6, 6, 10, 10, 10, 10, 10, 15, 15, 15, 15, 15, 15, 21, 21, 21, 21, 21, 21, 21, 28, 28, 28, 28, 28, 28, 28, 28, 36, 36, 36, 36, 36, 36, 36, 36, 36, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 66, 66, 66, 66, 66, 66

Offset: 0

Henry Bottomley, Oct 05 2000

			a(35) = 28 since 28 and 36 are successive triangular numbers and 28 <= 35 < 36.

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened

Cf. A000217, A003056, A056944, A057945, A127739.

a057944 n = a057944_list !! n          -- common flat access
a057944_list = concat a057944_tabl
a057944' n k = a057944_tabl !! n !! k  -- access when seen as a triangle
a057944_row n = a057944_tabl !! n
a057944_tabl = zipWith ($) (map replicate [1..]) a000217_list
-- Reinhard Zumkeller, Feb 03 2012

A057944 := proc(n)
        k := (-1+sqrt(1+8*n))/2 ;
        k := floor(k) ;
        k*(k+1)/2 ;
end proc; # R. J. Mathar, Nov 05 2011

f[n_] := Block[{a = Floor@ Sqrt[1 + 8 n]}, Floor[(a - 1)/2]*Floor[(a + 1)/2]/2]; Array[f, 72, 0]
t0=0; t1=1; k=1; Table[If[n < t1, t0, k++; t0=t1; t1=t1+k; t0], {n, 0, 72}]
With[{nn=15},Table[#[[1]],#[[2]]+1]&/@Thread[{Accumulate[Range[ 0,nn]],Range[ 0,nn]}]]//Flatten (* Harvey P. Dale, Mar 01 2020 *)

a(n)=my(t=(sqrtint(8*n+7)-1)\2);t*(t+1)/2 \\ Charles R Greathouse IV, Jan 26 2013

from math import comb, isqrt
def A057944(n): return comb((m:=isqrt(k:=n+1<<1))+(k>m*(m+1)),2) # Chai Wah Wu, Nov 09 2024

a(n) = floor((sqrt(1+8*n)-1)/2)*floor((sqrt(1+8*n)+1)/2)/2 = (trinv(n)*(trinv(n)-1))/2 = A000217(A003056(n)) = n - A002262(n)
a(n) = (1/2)*t*(t-1), where t = floor(sqrt(2*n+1)+1/2) = A002024(n+1). - Ridouane Oudra, Oct 20 2019
Sum_{n>=1} 1/a(n)^2 = 2*Pi^2/3 - 4. - Amiram Eldar, Aug 14 2022

Keyword tabl added by Reinhard Zumkeller, Feb 03 2012

A061336 Smallest number of triangular numbers which sum to n.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 2, 3, 1, 2, 3, 2, 3, 2, 1, 2, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 1, 2, 2, 2, 3, 3, 2, 2, 3, 1, 2, 3, 2, 2, 3, 2, 3, 3, 3, 1, 2, 2, 2, 3, 2, 2, 3, 3, 2, 2, 1, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 3, 1, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 3, 2, 1, 2, 2, 2, 3, 3, 2, 3, 2, 2, 2, 2, 3, 3

Offset: 0

Henry Bottomley, Apr 25 2001

nonn

a(n)=3 if n=5 or 8 mod 9, since triangular numbers are {0,1,3,6} mod 9.
From Bernard Schott, Jul 16 2022: (Start)
In September 1636, Fermat, in a letter to Mersenne, made the statement that every number is a sum of at most three triangular numbers. This was proved by Gauss, who noted this event in his diary on July 10 1796 with the notation:
EYPHKA! num = DELTA + DELTA + DELTA (where Y is in fact the Greek letter Upsilon and DELTA is the Greek letter of that name).
This proof was published in his book Disquisitiones Arithmeticae, Leipzig, 1801. (End)

			a(3)=1 since 3=3, a(4)=2 since 4=1+3, a(5)=3 since 5=1+1+3, with 1 and 3 being triangular.

Elena Deza and Michel Marie Deza, Fermat's polygonal number theorem, Figurate numbers, World Scientific Publishing (2012), Chapter 5, pp. 313-377.
C. F. Gauss, Disquisitiones Arithmeticae, Yale University Press, 1966, New Haven and London, p. 342, art. 293.

Giovanni Resta, Table of n, a(n) for n = 0..10000
George E. Andrews, EYPHKA! num = Delta + Delta + Delta, J. Number Theory 23 (1986), 285-293.
Eric Weisstein's World of Mathematics, Fermat's Polygonal Number Theorem.

Cf. A000217, A007294, A057945, A061337, A051533, A020757.

Cf. A100878 (analog for A000326), A104246 (analog for A000292), A283365 (analog for A000332), A283370 (analog for A000389).

t[n_]:=n*(n+1)/2; a[0]=0; a[n_]:=Block[ {k=1, tt= t/@ Range[Sqrt[2*n]]}, Off[IntegerPartitions::take]; While[{} == IntegerPartitions[n, {k}, tt, 1], k++]; k]; a/@ Range[0, 104] (* Giovanni Resta, Jun 09 2015 *)

\\ see A283370 for generic code, working but not optimized for this case of triangular numbers. - M. F. Hasler, Mar 06 2017

a(n)=my(m=n%9,f); if(m==5 || m==8, return(3)); f=factor(4*n+1); for(i=1,#f~, if(f[i,2]%2 && f[i,1]%4==3, return(3))); if(ispolygonal(n,3), n>0, 2) \\ Charles R Greathouse IV, Mar 17 2022

a(n) = 0 if n=0, otherwise 1 if n is in A000217, otherwise 2 if n is in A051533, otherwise 3 in which case n is in A020757.

a(n) <= 3 (proposed by Fermat and proved by Gauss). - Bernard Schott, Jul 16 2022

A006893 Smallest number whose representation requires n triangular numbers with greedy algorithm; also number of 1-2 rooted trees of height n.

Original entry on oeis.org

1, 2, 5, 20, 230, 26795, 359026205, 64449908476890320, 2076895351339769460477611370186680, 2156747150208372213435450937462082366919951682912789656986079991220

Offset: 1

Jeffrey Shallit

M. Abert and P. Diaconis, paper in preparation, 2002.
D. Parisse, The Tower of Hanoi and the Stern-Brocot-Array, Thesis, Munich, 1997.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Michael De Vlieger, Table of n, a(n) for n = 1..13
Mee Seong Im and Can Ozan Oğuz, Natural transformations between induction and restriction on iterated wreath product of symmetric group of order 2, (2021).
E. Lemoine, Note sur deux nouvelles décompositions des nombres entiers, Assoc. française pour l'avancement des sciences. Vol. 29, Tome 2, pp. 72-74, 1900.
Sridhar Narayanan, The Representation Theory of 2-Sylow Subgroups of the Symmetric Group, arXiv:1712.02507 [math.RT], 2017.
Index entries for sequences related to Stern's sequences
Index entries for sequences related to rooted trees

Where records occur in A057945, n >= 1.

Cf. A007501.

A006893 := proc(n) option remember; if n=1 then 1 else A006893(n-1)*(A006893(n-1)+3)/2; fi; end;

RecurrenceTable[{a[1] == 1, a[n] == a[n-1]*(a[n-1] + 3)/2}, a[n], {n, 10}] (* Vaclav Kotesovec, Dec 17 2014 *)

a=vector(20); a[1]=1; for(n=2, #a, a[n]=a[n-1]*(a[n-1]+3)/2); a \\ Altug Alkan, Apr 04 2018

a(n) = A007501(n-1) - 1.
a(n+1) = a(n)*(a(n)+3)/2, a(1)=1.
a(0) = 1, a(n) = Sum_{i=0..n-1} t(a(i)), where t(n)=n*(n+1)/2. - Jon Perry, Feb 14 2004
a(n) ~ 2 * c^(2^n), where c = 1.16007248510653786919452141287945841802404855231102953089... . - Vaclav Kotesovec, Dec 17 2014

Additional description from Andreas M. Hinz and Daniele Parisse

A000462 Numbers written in base of triangular numbers.

Original entry on oeis.org

1, 2, 10, 11, 12, 100, 101, 102, 110, 1000, 1001, 1002, 1010, 1011, 10000, 10001, 10002, 10010, 10011, 10012, 100000, 100001, 100002, 100010, 100011, 100012, 100100, 1000000, 1000001, 1000002, 1000010, 1000011, 1000012, 1000100, 1000101, 10000000, 10000001

Offset: 1

John Radu (Suttones(AT)aol.com)

A003056 and A057945 give lengths and sums. - Reinhard Zumkeller, Mar 27 2011

			The digits (from right to left) have values 1, 3, 6, 10, etc. (A000217), hence a(20) = 10012 because 20 = 1*15 + 0*10 + 0*6 + 1*3 + 2*1. - _Stefano Spezia_, Apr 25 2024

F. Smarandache, "Properties of the numbers", Univ. of Craiova Archives, 1975; Arizona State University Special Collections, Tempe, AZ.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
F. Iacobescu, Smarandache Partition Type and Other Sequences, Bull. Pure Appl. Sciences, Vol. 16E, No. 2 (1997), pp. 237-240.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
Eric Weisstein's World of Mathematics, Smarandache Sequences.

Cf. A000217, A003056, A057945.

a000462 n = g [] n $ reverse $ takeWhile (<= n) $ tail a000217_list where
   g as 0 []     = read $ concat $ map show $ reverse as :: Integer
   g as x (t:ts) = g (a:as) r ts where (a,r) = divMod x t
-- Reinhard Zumkeller, Mar 27 2011

A000217[n_]:=n(n+1)/2; a[n_]:=Module[{k=0}, num=n; digits={}; k=Floor[(Sqrt[1+8num]-1)/2]; While[num>0, AppendTo[digits, Floor[num/A000217[k]]]; num=Mod[num, A000217[k]]; kold=k; k=Floor[(Sqrt[1+8num]-1)/2]; While[kStefano Spezia, Apr 25 2024 *)

A280053 "Nachos" sequence based on squares.

Original entry on oeis.org

1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 3, 4, 2, 3, 4, 5, 6, 3

Offset: 1

N. J. A. Sloane, Jan 07 2017

nonn

The nachos sequence based on a sequence of positive numbers S starting with 1 is defined as follows: To find a(n) we start with a pile of n nachos.
During each phase, we successively remove S(1), then S(2), then S(3), ..., then S(i) nachos from the pile until fewer than S(i+1) remain. Then we start a new phase, successively removing S(1), then S(2), ..., then S(j) nachos from the pile until fewer than S(j+1) remain. Repeat. a(n) is the number of phases required to empty the pile.
Suggested by the Fibonachos sequence A280521, which is the case when S is 1,1,2,3,5,8,13,... (A000045).
If S = 1,2,3,4,5,... we get A057945.
If S = 1,2,3,5,7,11,... (A008578) we get A280055.
If S = triangular numbers we get A281367.
If S = squares we get the present sequence.
If S = powers of 2 we get A100661.
Needs a more professional Maple program.
Comment from Matthew C. Russell, Jan 30 2017 (Start):
Theorem: Any nachos sequence based on a sequence S = {1=s1 < s2 < s3 < ...} is unbounded.
Proof: S is the (infinite) set of numbers that we are allowed to subtract. (In the case of Fibonachos, this is the set of Fibonaccis themselves, not the partial sums.)
Suppose that n is a positive integer, with the number of stages of the process denoted by a(n).
Let s_m be the smallest element of S that is greater than n.
Then, if you start the process at N = n + s1 + s2 + s3 + ... + s_(m-1), you will get stuck when you hit n, and will have to start the process over again. Thus you will take a(n) + 1 stages of the process here, so a(N) = a(n) + 1.
(End)

			If n = 10, in the first phase we successively remove 1, then 4 nachos, leaving 5 in the pile. The next square is 9, which is bigger than 5, so we start a new phase. We remove 1, then 4 nachos, and now the pile is empty. There were two phases, so a(10)=2.

Lars Blomberg, Table of n, a(n) for n = 1..10000
Reddit user Teblefer, Fibonachos

Cf. A000045, A008578, A280521, A057945, A281367, A100661, A280055.

For indices of first occurrences of 1,2,3,4,... see A280054.

S:=[seq(i^2,i=1..1000)];
phases := proc(n) global S; local a,h,i,j,ipass;
a:=1; h:=n;
for ipass from 1 to 100 do
   for i from 1 to 100 do
      j:=S[i];
      if j>h then a:=a+1; break; fi;
      h:=h-j;
      if h=0 then return(a); fi;
                       od;
od;
return(-1);
end;
t1:=[seq(phases(i),i=1..1000)];
# 2nd program
A280053 := proc(n)
    local a,nres,i ;
    a := 0 ;
    nres := n;
    while nres > 0 do
        for i from 1 do
            if A000330(i) > nres then
                break;
            end if;
        end do:
        nres := nres-A000330(i-1) ;
        a := a+1 ;
    end do:
    a ;
end proc:
seq(A280053(n),n=1..80) ; # R. J. Mathar, Mar 05 2017

A280053[n_] := Module[{a, nres, i}, a = 0; nres = n; While[nres > 0, For[i = 1, True, i++, If[i(i+1)(2i+1)/6 > nres, Break[]]]; nres = nres - i(i-1)(2i-1)/6; a++]; a];
Table[A280053[n], {n, 1, 90}] (* Jean-François Alcover, Mar 16 2023, after R. J. Mathar *)

A281367 "Nachos" sequence based on triangular numbers.

Original entry on oeis.org

1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 2, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 2, 3, 4, 5, 3, 4, 5, 6, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 4

Offset: 1

N. J. A. Sloane, Jan 30 2017

nonn

The nachos sequence based on a sequence of positive numbers S starting with 1 is defined as follows: To find a(n) we start with a pile of n nachos.
During each phase, we successively remove S(1), then S(2), then S(3), ..., then S(i) nachos from the pile until fewer than S(i+1) remain. Then we start a new phase, successively removing S(1), then S(2), ..., then S(j) nachos from the pile until fewer than S(j+1) remain. Repeat. a(n) is the number of phases required to empty the pile.
Suggested by the Fibonachos sequence A280521, which is the case when S is 1,1,2,3,5,8,13,... (A000045).
If S = 1,2,3,4,5,... we get A057945.
If S = 1,2,3,5,7,11,... (A008578) we get A280055.
If S = triangular numbers we get the present sequence.
If S = squares we get A280053.
If S = powers of 2 we get A100661.
More than the usual number of terms are shown in order to distinguish this sequence from A104246.

			If n = 14, in the first phase we successively remove 1, then 3, then 6 nachos, leaving 4 in the pile. The next triangular number is 10, which is bigger than 4, so we start a new phase. We remove 1, then 3 nachos, and now the pile is empty. There were two phases, so a(14)=2.

Lars Blomberg, Table of n, a(n) for n = 1..10000
Reddit user Teblefer, Fibonachos

Cf. A000045, A008578, A280521, A057945, A100661, A280055.
For indices of first occurrences of 1,2,3,4,... see A281368.
Different from A104246.

S:=[seq(i*(i+1)/2,i=1..1000)];
phases := proc(n) global S; local a,h,i,j,ipass;
a:=1; h:=n;
for ipass from 1 to 100 do
for i from 1 to 100 do
j:=S[i];
if j>h then a:=a+1; break; fi;
h:=h-j;
if h=0 then return(a); fi;
od;
od;
return(-1);
end;
t1:=[seq(phases(i),i=1..1000)];
# 2nd program
A281367 := proc(n)
    local a,nres,i ;
    a := 0 ;
    nres := n;
    while nres > 0 do
        for i from 1 do
            if A000292(i) > nres then
                break;
            end if;
        end do:
        nres := nres-A000292(i-1) ;
        a := a+1 ;
    end do:
    a ;
end proc:
seq(A281367(n),n=1..80) ; # R. J. Mathar, Mar 05 2017

tri[n_] := n (n + 1) (n + 2)/6;
A281367[n_] := Module[{a = 0, nres = n, i}, While[nres > 0, For[i = 1, True, i++, If[tri[i] > nres, Break[]]]; nres -= tri[i-1]; a++]; a];
Table[A281367[n], {n, 1, 99}] (* Jean-François Alcover, Apr 11 2024, after R. J. Mathar *)

A120246 a(1) = 1. a(m(m+1)/2 + k) = m + a(k), 1 <= k <= m+1, m >= 1.

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 4, 5, 6, 6, 5, 6, 7, 7, 8, 6, 7, 8, 8, 9, 10, 7, 8, 9, 9, 10, 11, 10, 8, 9, 10, 10, 11, 12, 11, 12, 9, 10, 11, 11, 12, 13, 12, 13, 14, 10, 11, 12, 12, 13, 14, 13, 14, 15, 15, 11, 12, 13, 13, 14, 15, 14, 15, 16, 16, 15, 12, 13, 14, 14, 15, 16, 15, 16, 17, 17, 16, 17, 13

Offset: 1

Leroy Quet, Jun 12 2006

When calculating a(n) using the recurrence, the number of steps to reach a(1) is A057945(n-1). - Kevin Ryde, Aug 23 2022

Kevin Ryde, Table of n, a(n) for n = 1..10011
Kevin Ryde, PARI/GP Code

Cf. A120244, A120245, A057945.

PARI
```
\\ See links.
```

Extended by Ray Chandler, Jun 19 2006

A364885 Triangle T(n, k), n >= 0, k = 0..n, read by rows; T(0, 0) = 0, and for any n > 0, k = 0..n, T(n, k) is the least number obtained by turning a 0 into a 1 in the binary expansion of the k-th term of the (0-based) flattened sequence.

Original entry on oeis.org

0, 1, 3, 2, 5, 7, 4, 9, 11, 6, 8, 17, 19, 10, 13, 16, 33, 35, 18, 21, 15, 32, 65, 67, 34, 37, 23, 12, 64, 129, 131, 66, 69, 39, 20, 25, 128, 257, 259, 130, 133, 71, 36, 41, 27, 256, 513, 515, 258, 261, 135, 68, 73, 43, 14, 512, 1025, 1027, 514, 517, 263, 132, 137, 75, 22, 24

Offset: 0

Rémy Sigrist, Aug 12 2023

In other words, T(n, k) = a(k) OR 2^e for some e >= 0 (where OR denotes the bitwise OR operator).

As a flat sequence, this is a permutation of the nonnegative integers (as, for any h >= 0, the sequence contains all numbers with Hamming weight h); see A365080 for the inverse.

			Triangle begins:
    0
    1, 3
    2, 5, 7
    4, 9, 11, 6
    8, 17, 19, 10, 13
    16, 33, 35, 18, 21, 15
    32, 65, 67, 34, 37, 23, 12
    64, 129, 131, 66, 69, 39, 20, 25
    128, 257, 259, 130, 133, 71, 36, 41, 27
    256, 513, 515, 258, 261, 135, 68, 73, 43, 14
    512, 1025, 1027, 514, 517, 263, 132, 137, 75, 22, 24
    ...

Rémy Sigrist, PARI program
Index entries for sequences that are permutations of the natural numbers

See A364884 for a similar sequence.

Cf. A000120, A057945, A365080 (inverse).

PARI
```
See Links section.
```

T(n, 0) = 2^(n-1) for any n > 0.

A000120(a(n)) = A057945(n).

A145172 Number of pentagonal numbers needed to represent n with greedy algorithm.

Original entry on oeis.org

1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 2, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 2, 3, 4, 5, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 2, 3, 4, 5, 6, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 2, 3, 4, 5, 6, 3, 4, 5, 6

Offset: 1

Christina Steffan (christina.steffan(AT)gmx.at), Oct 03 2008

nonn

Sequence is unbounded.

			a(21)=6 since 21 = 12+5+1+1+1+1.

Andrew Howroyd, Table of n, a(n) for n = 1..10000

Cf. A000326 (pentagonal numbers), A053610, A057945, A180447, A192988.

a(n)={my(s=0); forstep(k=(sqrtint(24*n+1)+1)\6, 1, -1, my(t=k*(3*k-1)/2); s+=n\t; n%=t); s} \\ Andrew Howroyd, Apr 21 2021

Terms a(41) and beyond from Andrew Howroyd, Apr 21 2021

cp's OEIS Frontend

A053610 Number of positive squares needed to sum to n using the greedy algorithm.

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Formula

A057944 Largest triangular number less than or equal to n; write m-th triangular number m+1 times.

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Haskell

Maple

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PARI

Python

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A061336 Smallest number of triangular numbers which sum to n.

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Mathematica

PARI

PARI

Formula

A006893 Smallest number whose representation requires n triangular numbers with greedy algorithm; also number of 1-2 rooted trees of height n.

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Maple

Mathematica

PARI

Formula

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A000462 Numbers written in base of triangular numbers.

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Haskell

Mathematica

A280053 "Nachos" sequence based on squares.

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Maple