A183532 -id:A183532 - cp's OEIS frontend

A184535 a(n) = floor(3/5 * n^2), with a(1)=1.

1, 2, 5, 9, 15, 21, 29, 38, 48, 60, 72, 86, 101, 117, 135, 153, 173, 194, 216, 240, 264, 290, 317, 345, 375, 405, 437, 470, 504, 540, 576, 614, 653, 693, 735, 777, 821, 866, 912, 960, 1008, 1058, 1109, 1161, 1215, 1269, 1325, 1382, 1440, 1500, 1560, 1622, 1685, 1749, 1815, 1881, 1949, 2018, 2088, 2160, 2232, 2306, 2381

Offset: 1

Clark Kimberling, Jan 16 2011

Apart from the initial term this is the elliptic troublemaker sequence R_n(2,5) in the notation of Stange (see Table 1, p.16). For other elliptic troublemaker sequences see the cross references below. - Peter Bala, Aug 08 2013

Ray Chandler, Table of n, a(n) for n = 1..10000
K. E. Stange, Integral points on elliptic curves and explicit valuations of division polynomials arXiv:1108.3051v3 [math.NT]
Wikipedia, Maximum number of deciamonds in a hexagon.
Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 0, 1, -2, 1).

Cf. A183532, A279169.

Elliptic troublemaker sequences: A000212 (= R_n(1,3) = R_n(2,3)), A002620 (= R_n(1,2)), A007590 (= R_n(2,4)), A030511 (= R_n(2,6) = R_n(4,6)), A033436 (= R_n(1,4) = R_n(3,4)), A033437 (= R_n(1,5) = R_n(4,5)), A033438 (= R_n(1,6) = R_n(5,6)), A184535 (= R_n(2,5) = R_n(3,5)).

Concatenation([1], List([2..10^3], n->Int(3/5 * n^2))); # Muniru A Asiru, Feb 04 2018

1,seq(floor(3/5*n^2), n=2..10^3); # Muniru A Asiru, Feb 04 2018

p[n_] := FractionalPart[(n^3 + 5)^(1/3)]; q[n_] := Floor[1/p[n]]; Table[q[n], {n, 1, 120}]
Join[{1},LinearRecurrence[{2, -1, 0, 0, 1, -2, 1},{2, 5, 9, 15, 21, 29, 38},62]] (* Ray Chandler, Aug 31 2015 *)

a(n) = if(n==1, 1, 3*n^2\5); \\ Altug Alkan, Mar 03 2018

def A184535(n): return 3*n**2//5 if n>1 else 1 # Chai Wah Wu, Aug 04 2025

a(n) = floor(1/{(5+n^3)^(1/3)}), where {}=fractional part.
a(n)= +2*a(n-1) -a(n-2) +a(n-5) -2*a(n-6) +a(n-7), for n>8, with g.f. 1-x^2*(1+x)*(2*x^2-x+2)/ ((x^4+x^3+x^2+x+1) *(x-1)^3), so a(n) is (3n^2-2)/5 plus a fifth of A164116 for n>1. [Bruno Berselli, Jan 30 2011. See the following Bala's comment.]
From Peter Bala, Aug 08 2013: (Start)
a(n) = floor(3/5*n^2) for n >= 2.
The sequence b(n) := floor(3/5*n^2) - 3/5*n^2, n >= 1, is periodic with period [-3/5, -2/5, -2/5, -3/5, 0] of length 5. The generating function and recurrence equation given above easily follow from these observations.
The sequence c(n) := 5/2*( (2*n/5 - floor(2*n/5))^2 - (2*n/5 - floor(2*n/5)) ) is also periodic with period 5, and calculation shows it has the same period as the sequence b(n). Thus b(n) = c(n), yielding the alternative formula a(n) = 3/5*n^2 + 5/2*( (2*n/5 - floor(2*n/5))^2 - (2*n/5 - floor(2*n/5)) ), which is one of the formulas for the elliptic troublemaker sequence R_n(2,5) given in Stange (see Section 7, equation (21)). (End)

Better name from Peter Bala, Aug 08 2013

A183534 Square array of generalized Ulam numbers U(n,k), n>=1, k>=2, read by antidiagonals: U(n,k) = n if n<=k; for n>k, U(n,k) = least number > U(n-1,k) which is a unique sum of k distinct terms U(i,k) with i

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 6, 6, 1, 2, 3, 4, 9, 8, 1, 2, 3, 4, 10, 10, 11, 1, 2, 3, 4, 5, 16, 11, 13, 1, 2, 3, 4, 5, 15, 17, 12, 16, 1, 2, 3, 4, 5, 6, 25, 18, 28, 18, 1, 2, 3, 4, 5, 6, 21, 26, 19, 29, 26, 1, 2, 3, 4, 5, 6, 7, 36, 27, 22, 30, 28, 1, 2, 3, 4, 5, 6, 7, 28, 37, 28, 64, 53, 36

Offset: 1

Jonathan Vos Post and Alois P. Heinz, Jan 05 2011

The columns are Ulam-type sequences - see A002858 for further information.  Some of these sequences - but not all - seem to have quite simple generating functions.
U(k+1,k)   = k*(k+1)/2.
U(k+2+j,k) = k^2+j for k>=3 and 0<=jU(2*k+2,k) = k*(3*k-1)/2 for k>=3.

			Square array U(n,k) begins:
  1,  1,  1,  1,  1,  1,  ...
  2,  2,  2,  2,  2,  2,  ...
  3,  3,  3,  3,  3,  3,  ...
  4,  6,  4,  4,  4,  4,  ...
  6,  9, 10,  5,  5,  5,  ...
  8, 10, 16, 15,  6,  6,  ...

Alois P. Heinz, Antidiagonals n = 1..87
Index entries for Ulam numbers

Columns k=2-10 give: A002858, A007086, A183527, A183528, A183529, A183530, A183531, A183532, A183533.

Cf. A135737.

b:= proc(n,i,k,h) option remember;
      local t;
      if n<0 or h<0 then 0
    elif n=0 then `if`(h=0, 1, 0)
    elif i=0 or h=0 then 0
    elif h=1 then t:= v(n, k);
                  `if`(t>0 and t<=i, 1, 0)
             else t:= b(n -U(i, k), i-1, k, h-1);
                  t+ `if`(t>1, 0, b(n, i-1, k, h))
      fi
    end:
v:= proc() 0 end:
U:= proc(n,k) option remember;
      local m;
      if n<=k then v(n,k):= n
              else for m from U(n-1, k)+1
                     while b(m, n-1, k, k)<>1 do od;
                   v(m,k):= n; m
      fi
    end:
seq(seq(U(n, 2+d-n), n=1..d), d=1..12);

b[n_, i_, k_, h_] := b[n, i, k, h] = Module[{t}, Which[n < 0 || h < 0, 0, n == 0, If[h == 0, 1, 0], i == 0 || h == 0, 0, h == 1, t = v[n, k]; If[t > 0 && t <= i, 1, 0], True, t = b[n-U[i, k], i-1, k, h-1]; t+If[t > 1, 0, b[n, i-1, k, h]] ] ]; v[, ] = 0; U[n_, k_] := U[n, k] = Module[{m}, If[n <= k, v[n, k] = n, For[m = U[n-1, k]+1, b[m, n-1, k, k] != 1, m++]; v[m, k] = n; m] ]; Table[Table[U[n, 2+d-n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Dec 23 2013, translated from Maple *)

Ulam(N,k=2,v=0)={ my( a=vector(k,i,i), c );
for( n=k,N-1, for( t=1+a[n],9e9, c=0;
  forvec(v=vector(k,i,[i,n]),sum(j=1,k,a[v[j]])==t & c++>1 & next(2),2);
  c||next; v&print1(t","); a=concat(a,t); break;
)); a}
/* M. F. Hasler */

A183533 An Ulam-type sequence: a(n) = n if n<=10; for n>10, a(n) = least number > a(n-1) which is a unique sum of 10 distinct earlier terms.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 55, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 145, 163, 190, 217, 235, 271, 280, 1740, 1741, 1744, 1745, 1799, 1804, 1805, 1824, 1825, 1831, 1859, 1869, 1913, 1914, 3554, 10521, 10522, 10526, 10527, 10537, 10563, 10564

Offset: 1

Jonathan Vos Post and Alois P. Heinz, Jan 05 2011

nonn

An Ulam-type sequence - see A002858 for further information.

			a(11) = 55 = 1 + ... + 10 = 10*11/2, because it is the least number >10 with a unique sum of 10 distinct earlier terms.
a(12) = 100 = 1 + ... + 9 + 55 = 10^2, because it is the least number >55 with a unique sum of 10 distinct earlier terms.

Alois P. Heinz, Table of n, a(n) for n = 1..120
Index entries for Ulam numbers

Column k=10 of A183534.

Cf. A002858, A007086, A183527-A183532, A135737.

Maple
```
# see A183534 for programs.
```

A184533 a(n) = floor(1/{(2+n^3)^(1/3)}), where {}=fractional part.

Original entry on oeis.org

2, 6, 13, 24, 37, 54, 73, 96, 121, 150, 181, 216, 253, 294, 337, 384, 433, 486, 541, 600, 661, 726, 793, 864, 937, 1014, 1093, 1176, 1261, 1350, 1441, 1536, 1633, 1734, 1837, 1944, 2053, 2166, 2281, 2400, 2521, 2646, 2773, 2904, 3037, 3174, 3313, 3456, 3601

Offset: 1

Clark Kimberling, Jan 16 2011

Column 2 of the array at A184532.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A183532, A183534. Essenitally the same as A032528.

p[n_]:=FractionalPart[(n^3+2)^(1/3)]; q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,120}]
Join[{2},Table[(6*n^2 - (1-(-1)^n))/4,{n,2,50}]] (* or *) Join[{2}, LinearRecurrence[{2,0,-2,1},{6, 13, 24, 37},50]] (* G. C. Greubel, Feb 20 2017 *)

a(n)=my(x=sqrtn(n^3+2,3));x-=n;1/x\1 \\ Charles R Greathouse IV, Aug 23 2011

concat([2], for(n=2,25, print1((6*n^2 - (1-(-1)^n))/4, ", "))) \\ G. C. Greubel, Feb 20 2017

a(n) = floor(1/{(2+n^3)^(1/3)}), where {}=fractional part.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(n) = (6*n^2 - (1-(-1)^n))/4 for n>1.
From Alexander R. Povolotsky, Aug 22 2011: (Start)
a(n+1) +a(n) = 3*n^2 + 3*n + 1.
G.f.: x*(-2 - 2*x - x^2 - 2*x^3 + x^4)/((-1 + x)^3*(1 + x)). (End)
a(n) = floor(1/((n^3+2)^(1/3)-n)). - Charles R Greathouse IV, Aug 23 2011
E.g.f.: (3*x*(x + 1)*cosh(x) + (3*x^2 + 3*x - 1)*sinh(x) + 2*x)/2. - Stefano Spezia, Apr 19 2025

A184534 a(n) = floor(1/{(4+n^3)^(1/3)}), where {}=fractional part.

Original entry on oeis.org

1, 3, 7, 12, 18, 27, 36, 48, 60, 75, 90, 108, 126, 147, 168, 192, 216, 243, 270, 300, 330, 363, 396, 432, 468, 507, 546, 588, 630, 675, 720, 768, 816, 867, 918, 972, 1026, 1083, 1140, 1200, 1260, 1323, 1386, 1452, 1518, 1587, 1656, 1728, 1800, 1875, 1950, 2028, 2106, 2187, 2268, 2352, 2436, 2523, 2610, 2700, 2790, 2883, 2976, 3072, 3168, 3267, 3366, 3468, 3570, 3675, 3780, 3888, 3996, 4107, 4218, 4332, 4446, 4563, 4680, 4800, 4920, 5043, 5166, 5292, 5418, 5547, 5676, 5808

Offset: 1

Clark Kimberling, Jan 16 2011

nonn

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A183532, A183534.

Table[Floor[1/FractionalPart[(n^3 + 4)^(1/3)]], {n, 1, 120}]

for(n=1, 50, print1(floor(1/frac((4 + n^3)^(1/3))), ", ")) \\ G. C. Greubel, May 14 2017

a(n) = floor[1/{(4+n^3)^(1/3)}], where {}=fractional part.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
From Colin Barker, Oct 07 2012: (Start)
Empirical: a(n) = 3*(1 - (-1)^n + 4*n + 2*n^2)/8 for n>2.
Empirical G.f.: x*(x^6-2*x^5+x^4-x^2-x-1)/((x-1)^3*(x+1)).(End)

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A184535 a(n) = floor(3/5 * n^2), with a(1)=1.

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A183534 Square array of generalized Ulam numbers U(n,k), n>=1, k>=2, read by antidiagonals: U(n,k) = n if n<=k; for n>k, U(n,k) = least number > U(n-1,k) which is a unique sum of k distinct terms U(i,k) with i

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A183533 An Ulam-type sequence: a(n) = n if n<=10; for n>10, a(n) = least number > a(n-1) which is a unique sum of 10 distinct earlier terms.

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Maple

A184533 a(n) = floor(1/{(2+n^3)^(1/3)}), where {}=fractional part.

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A184534 a(n) = floor(1/{(4+n^3)^(1/3)}), where {}=fractional part.

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