A001521 -id:A001521 - cp's OEIS frontend

A241576 Third differences of A001521.

0, 1, 0, 0, 1, 0, 1, 2, 2, 2, 4, 6, 8, 10, 16, 22, 31, 44, 62, 88, 124, 176, 248, 352, 497, 702, 994, 1406, 1987, 2812, 3975, 5622, 7951, 11244, 15902, 22488, 31804, 44976, 63607, 89954, 127213, 179908, 254427, 359814, 508855, 719628, 1017709

Offset: 1

N. J. A. Sloane, Apr 27 2014

nonn

Cf. A001521, A017911, A190660.

Differences[NestList[Floor[Sqrt[2#(#+1)]]&,1,50],3] (* Harvey P. Dale, Dec 17 2019 *)

A017911 Powers of sqrt(2) rounded to nearest integer.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 8, 11, 16, 23, 32, 45, 64, 91, 128, 181, 256, 362, 512, 724, 1024, 1448, 2048, 2896, 4096, 5793, 8192, 11585, 16384, 23170, 32768, 46341, 65536, 92682, 131072, 185364, 262144, 370728, 524288

Offset: 0

N. J. A. Sloane

Apart from offset the same as A057048. - T. D. Noe, Apr 27 2003
Indeed, write the natural numbers as triangle, [1; 2, 3; 4, 5, 6; ...], then the last number in each row is T(n) = n(n+1)/2 = A000217(n), and 2^k is located in the row n with n(n-1)/2 < 2^k <= n(n+1)/2 <=> n^2 - n < 2^(k+1) <= n^2 + n, which means that n = round(sqrt(2^(k+1))). - M. F. Hasler, Feb 20 2012
The rounded curvature of circle in square inscribing or the rounded radius of circle in square circumscribing with initial circle radius = 1 for both cases, see illustration in link. - Kival Ngaokrajang, Aug 07 2013
Even-indexed terms are powers of 2.

			sqrt(2)^3 = 2.82842712474619..., so a(3) = 3.
sqrt(2)^4 = 4, so a(4) = 4.
sqrt(2)^5 = 5.6568542494923801952..., so a(5) = 6.
sqrt(2)^6 = 8, so a(6) = 8.
sqrt(2)^7 = 11.31370849898476..., so a(7) = 11.

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Kival Ngaokrajang, Illustration of initial terms of square inscribing
Eric Weisstein's World of Mathematics, Polygon Inscribing

Cf. A000079, A057048.

Apart from offset, first differences of A001521.

[Round(Sqrt(2)^n): n in [0..40]]; // Vincenzo Librandi, Nov 19 2011

Floor[(Sqrt[2]^Range[0, 40] + 1/2)] (* Vincenzo Librandi, Nov 19 2011 *)

a(n)=round(sqrt(2)^n) \\ Charles R Greathouse IV, Nov 18 2011

from math import isqrt
def A017911(n): return -isqrt(m:=1<Chai Wah Wu, Jun 18 2024

A091523 Graham-Pollak sequence with initial term 8.

Original entry on oeis.org

8, 12, 17, 24, 34, 48, 68, 96, 136, 193, 273, 386, 546, 772, 1092, 1545, 2185, 3090, 4370, 6180, 8740, 12360, 17480, 24721, 34961, 49443, 69923, 98886, 139846, 197772, 279692, 395544, 559384, 791089, 1118769, 1582179, 2237539, 3164358

Offset: 1

Eric W. Weisstein, Jan 18 2004

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..500
Th. Stoll, On Families of Nonlinear Recurrences Related to Digits, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.2.
Eric Weisstein's World of Mathematics, Graham-Pollak Sequence

Cf. A001521, A091522.

See A001521.

A091522 Graham-Pollak sequence with initial term 5.

Original entry on oeis.org

5, 7, 10, 14, 20, 28, 40, 57, 81, 115, 163, 231, 327, 463, 655, 927, 1311, 1854, 2622, 3708, 5244, 7416, 10488, 14832, 20976, 29665, 41953, 59331, 83907, 118663, 167815, 237326, 335630, 474653, 671261, 949307, 1342523, 1898614, 2685046

Offset: 1

Eric W. Weisstein, Jan 18 2004

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..500
Th. Stoll, On Families of Nonlinear Recurrences Related to Digits, Journal of Integer Sequences, Vol. 8 (2005), Article 05.3.2.
Eric Weisstein's World of Mathematics, Graham-Pollak Sequence

Cf. A001521, A091523.

NestList[Floor[Sqrt[2](#+1/2)]&,5,40] (* Harvey P. Dale, Feb 24 2018 *)

first(n)=my(v=vector(n)); v[1]=5; for(k=2,n, v[k]=sqrtint(2*(v[k-1]+1)*v[k-1])); v \\ Charles R Greathouse IV, Jan 23 2020

a(n) = floor(sqrt(2) * (a(n-1) + 1/2)).

A190660 Number of triangular numbers T(k) between powers of 2, 2^(n-1) < T(k) <= 2^n.

Original entry on oeis.org

1, 0, 1, 1, 2, 2, 3, 5, 7, 9, 13, 19, 27, 37, 53, 75, 106, 150, 212, 300, 424, 600, 848, 1200, 1697, 2399, 3393, 4799, 6786, 9598, 13573, 19195, 27146, 38390, 54292, 76780, 108584, 153560, 217167, 307121, 434334, 614242, 868669, 1228483, 1737338, 2456966

Offset: 0

John W. Nicholson, May 16 2011

nonn

Count of triangular numbers between powers of 2.
a(n)/a(n-1) converges to sqrt(2) (A002193). [John W. Nicholson, May 16 2011]
Essentially first differences of A017911. - Jeremy Gardiner, Aug 11 2013. Also second differences of A001521. - N. J. A. Sloane, Apr 27 2014

			Between 2^(6-1)=32 and 2^6=64 are T(8)=36, T(9)=45, T(10)=55 so A190660(6)=3.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A001521, A002193, A017911.

TriangularIndex[n_] := (-1 + Sqrt[1 + 8*n])/2; Differences[Table[Floor[TriangularIndex[2^n]], {n, -1, 50}]] (* T. D. Noe, May 19 2011 *)

a(n) = if (n==0, 1, sum(i=2^(n-1)+1, 2^n, ispolygonal(i, 3))); \\ Michel Marcus, Apr 28 2014

Extended by T. D. Noe, May 19 2011

A383032 Exponent of the highest power of 2 dividing the n-th number that is either a square or twice a square.

Original entry on oeis.org

0, 1, 2, 3, 0, 4, 1, 0, 5, 2, 0, 1, 6, 3, 0, 1, 2, 0, 7, 4, 1, 0, 2, 3, 0, 1, 8, 5, 0, 2, 1, 0, 3, 4, 0, 1, 2, 9, 0, 6, 1, 0, 3, 2, 1, 0, 4, 5, 0, 1, 2, 0, 3, 10, 1, 0, 7, 2, 0, 1, 4, 3, 0, 2, 1, 0, 5, 6, 0, 1, 2, 3, 0, 1, 4, 0, 11, 2, 1, 0, 8, 3, 0, 1, 2, 5, 0

Offset: 1

Amiram Eldar, Apr 13 2025

The first position of k in this sequence, for k >= 0, is A001521(k+1).

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001521, A001954, A003151, A003152, A007814, A014176, A028982, A215247, A342280.

With[{m = 3000}, IntegerExponent[Union[Join[Range[Floor[Sqrt[m]]]^2, 2*Range[Floor[Sqrt[m/2]]]^2]], 2]]

lista(nn) = apply(x->valuation(x,2), vecsort(concat(vector(sqrtint(nn\1), i, i^2), vector(sqrtint(nn\2), i, 2*i^2)))); \\ Michel Marcus, Apr 13 2025

a(n) = A007814(A028982(n)).
a(A001954(n)) = 0 for n >= 0.
a(A215247(n)) = 1 for n >= 1.
a(A342280(n)) = 2 for n >= 0.
Sum_{k=1..n} a(k) ~ (1 + sqrt(2)) * n.

A091524 a(m) is the multiplier of sqrt(2) in the constant alpha(m) = a(m)*sqrt(2) - b(m), where alpha(m) is the value of the constant determined by the binary bits in the recurrence associated with the Graham-Pollak sequence.

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 3, 5, 4, 6, 7, 5, 8, 6, 9, 7, 10, 11, 8, 12, 9, 13, 14, 10, 15, 11, 16, 12, 17, 18, 13, 19, 14, 20, 21, 15, 22, 16, 23, 24, 17, 25, 18, 26, 19, 27, 28, 20, 29, 21, 30, 31, 22, 32, 23, 33, 24, 34, 35, 25, 36, 26, 37, 38, 27, 39, 28, 40, 41, 29, 42, 30, 43, 31, 44

Offset: 1

Eric W. Weisstein, Jan 18 2004

nonn

Each integer appears twice. If one deletes the first occurrence of each positive integer one obtains the sequence of positive integers: 1,2,3,4,5,...; i.e., if we enclose in parentheses the first occurrence of 1,2,3,... giving (1),1,(2),2,(3),(4),3,(5),4,(6),(7),5,(8),6,(9),7,(10),... and remove them, we obtain: 1,2,3,4,5,6,7,... The same property holds if one deletes the second occurrence of each positive integer. - Benoit Cloitre, Oct 13 2007

			-1+sqrt(2), -1+sqrt(2), -2+2*sqrt(2), -2+2*sqrt(2), -4+3*sqrt(2), ..., so the sequence of multipliers is 1, 1, 2, 2, 3, ...

Eric Weisstein's World of Mathematics, Graham-Pollak Sequence

Cf. A001521, A003151, A003152, A006337.

Sequence is completely defined by: a(floor(n*(1+sqrt(2))))=n; a(floor(n*(1+1/sqrt(2))))=n, n>=1 since A003151 and A003152 are Beatty sequences partitioning the integers. - Benoit Cloitre, Oct 13 2007

Conjecture: a(n) = sqrt(A028982(n)/A006337(n)). - Mikhail Kurkov, Apr 25 2024

A100673 A Graham-Pollak-like sequence with cube root instead of square root.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 15, 20, 26, 34, 44, 56, 71, 90, 114, 144, 182, 230, 291, 367, 463, 584, 737, 929, 1171, 1476, 1860, 2344, 2954, 3723, 4691, 5911, 7448, 9385, 11825, 14899, 18772, 23652, 29800, 37546, 47306, 59603, 75096, 94616, 119209, 150195, 189235

Offset: 0

Jonathan Vos Post, Dec 06 2004

When the multiplier in the recurrence is 2 and the recurrence has two terms inside a square root, we have the Graham-Pollak sequence, where there is a remarkable exact explicit formula for a(n) in terms of the union of the set of integers and the set of integer multiples of Sqrt(2). As Weisstein summarizes Borwein & Bailey: "It is not known if sequences such as ... a(n) = a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3)) have corresponding properties." This sequence is the given one, having the multiplier in the recurrence as 2 and three terms inside a cube root and with a(0) = 1. Through n=50, the primes are when n = 1, 2, 3, 6, 13, 20, 21, 24, 25, 31. Through n=50, the semiprimes are when n = 4, 7, 9, 10, 19, 23, 32, 34, 36, 40, 42, 47, 49.

			a(6) = 11 because a(5) = 8; so a(6) = Floor((2*8*(8+1)*(8+2))^(1/3))
= floor(1440^(1/3)) = 11, which happens to be prime.
a(45) = 119209 because a(44) = 94616, so a(45) = Floor((2*94616*(94616+1)*(94616+2))^(1/3)) = floor(1694094050176992^(1/3)) = 119209 = 23 * 71 * 73, which happens to be a 3-brilliant number.

Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.
Eric Weisstein's World of Mathematics, Graham-Pollak sequence

Cf. A001521, A091522, A091523.

NestList[Floor[Surd[2#(#+1)(#+2),3]]&,1,50] (* Harvey P. Dale, Feb 24 2016 *)

a(0) = 1, a(n) = Floor((2*a(n-1)*(a(n-1)+1)*(a(n-1)+2))^(1/3))

A228560 Curvature (rounded down) of the circle inscribed in the n-th golden triangle arranged in a spiral form.

Original entry on oeis.org

2, 4, 7, 11, 18, 30, 49, 79, 129, 209, 338, 547, 886, 1434, 2320, 3754, 6075, 9830, 15905, 25735, 41641, 67376, 109017, 176394, 285412, 461806, 747218, 1209024

Offset: 1

Kival Ngaokrajang, Aug 25 2013

Starting with a golden triangle whose base is of length 1 and whose sides are of length phi = (1+sqrt(5))/2, create the next golden triangle at the base of the previous triangle, i.e., sides length = 1 and base length = phi-1, and so on. a(n) is the floor of the curvature (inverse of the radius) of the circle inscribed in the n-th triangle.
The golden triangles created by this process are the same as the golden triangles inscribed in a logarithmic spiral.
The logarithmic spiral can be approximated by circular arcs of radii 1, phi-1, (phi-1)^2, ... which are the sides of bisected golden gnomons and center located at their related apex. The sequence whose n-th term is the curvature (rounded down) of the n-th such circular arc is A014217. See illustration in link.

Kival Ngaokrajang, Illustration of initial terms.
Wikipedia, Golden triangle.

Cf. A001521 (for 45-45-90 triangles), A065565 (for 3:4:5 triangles), A014217.

A100671 A Graham-Pollak-like sequence with multiplier 3 instead of 2.

Original entry on oeis.org

1, 2, 4, 7, 12, 21, 37, 64, 111, 193, 335, 581, 1007, 1745, 3023, 5236, 9069, 15708, 27207, 47124, 81622, 141374, 244867, 424122, 734601, 1272367, 2203805, 3817103, 6611417, 11451311, 19834253, 34353934, 59502759, 103061802, 178508278, 309185407, 535524834

Offset: 0

Jonathan Vos Post, Dec 06 2004

nonn

When the multiplier in the recurrence is 2, we have the Graham-Pollak sequence, where there is a remarkable exact explicit formula for a(n) in terms of the union of the set of integers and the set of integer multiples of Sqrt(2). As Weisstein summarizes Borwein & Bailey: "It is not known if sequences such as a(n) = Floor(Sqrt(3*a(n-1)*(a(n-1)+1))) have corresponding properties." This sequence is the given one, with a(0) = 1. Through n=40, the primes are when n = 1, 3, 6, 9, 14, 25, 28, 29. Through n=40, the semiprimes are when n = 2, 5, 8, 10, 11, 12, 13, 16, 21, 22, 26, 27, 30, 31, 38.

			a(9) = 193 because a(8) = 111; so a(9) = Floor(Sqrt(3*111*(111+1))) = floor(sqrt(37296)) = 193, which happens to be prime.

Borwein, J. and Bailey, D., Mathematics by Experiment: Plausible Reasoning in the 21st Century. Natick, MA: A. K. Peters, 2003.

R. L. Graham and H. O. Pollak, Note on a nonlinear recurrence related to sqrt(2), Mathematics Magazine, Volume 43, Pages 143-145, 1970. Zbl 201.04705.
Eric Weisstein's World of Mathematics, Graham-Pollak sequence

Cf. A001521, A091522, A091523.

RecurrenceTable[{a[0]==1,a[n]==Floor[Sqrt[3a[n-1](a[n-1]+1)]]},a[n],{n,40}] (* Harvey P. Dale, Sep 10 2011 *)

a(0) = 1, a(n) = Floor(Sqrt(3*a(n-1)*(a(n-1)+1))).

Corrected and extended by Harvey P. Dale, Sep 10 2011

cp's OEIS Frontend

A241576 Third differences of A001521.

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A017911 Powers of sqrt(2) rounded to nearest integer.

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A091523 Graham-Pollak sequence with initial term 8.

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A091522 Graham-Pollak sequence with initial term 5.

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A190660 Number of triangular numbers T(k) between powers of 2, 2^(n-1) < T(k) <= 2^n.

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A383032 Exponent of the highest power of 2 dividing the n-th number that is either a square or twice a square.

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A091524 a(m) is the multiplier of sqrt(2) in the constant alpha(m) = a(m)*sqrt(2) - b(m), where alpha(m) is the value of the constant determined by the binary bits in the recurrence associated with the Graham-Pollak sequence.

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A100673 A Graham-Pollak-like sequence with cube root instead of square root.

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