A309779 -id:A309779 - cp's OEIS frontend

A153894 a(n) = 5*2^n - 1.

4, 9, 19, 39, 79, 159, 319, 639, 1279, 2559, 5119, 10239, 20479, 40959, 81919, 163839, 327679, 655359, 1310719, 2621439, 5242879, 10485759, 20971519, 41943039, 83886079, 167772159, 335544319, 671088639, 1342177279, 2684354559

Offset: 0

Vladimir Joseph Stephan Orlovsky, Jan 03 2009

a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n + 1, where the string "{}" is used to represent the empty set and spaces are ignored. - Ely Golden, Nov 14 2019

a(n) converted to binary is 100 followed by n ones. - Alexandre Herrera, Oct 06 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Bernard Monjardet, Acyclic domains of linear orders: a survey, in "The Mathematics of Preference, Choice and Order: Essays in Honor of Peter Fishburn", edited by Steven Brams, William V. Gehrlein and Fred S. Roberts, Springer, 2009, pp. 139-160. This version: . - _N. J. A. Sloane_, Feb 07 2009
Gennady Eremin, Partitioning the set of natural numbers into Mersenne trees and into arithmetic progressions; Natural Matrix and Linnik's constant, arXiv:2405.16143 [math.CO], 2024. See pp. 3-5, 14.
Gennady Eremin, Infinite matrix of odd natural numbers. A bit about Sophie Germain prime numbers, arXiv:2501.17090 [math.GM], 2025. See pp. 3, 11.
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A052549, A309779.

[5*2^n-1: n in [0..30]]; // Vincenzo Librandi, Oct 28 2011

a=4;lst={a};Do[a=a*2+1;AppendTo[lst,a],{n,5!}];lst
LinearRecurrence[{3,-2},{4,9}, 25] (* or *) Table[5*2^n - 1, {n,0,25}] (* G. C. Greubel, Sep 01 2016 *)

a(n)=5*2^n-1 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 2*a(n-1) + 1, n>0.
a(n) = A052549(n+1).
G.f.: (4 - 3*x) / ( (2*x-1)*(x-1) ). - R. J. Mathar, Oct 22 2011
a(n) + a(n-1)^2 = A309779(n), a perfect square. - Vincenzo Librandi, Oct 28 2011
From G. C. Greubel, Sep 01 2016: (Start)
a(n) = 3*a(n-1) - 2*a(n-2).
E.g.f.: 5*exp(2*x) - exp(x). (End)

Edited by N. J. A. Sloane, Feb 07 2009

Definition corrected by Franklin T. Adams-Watters, Apr 22 2009

A309778 a(n) is the greatest integer such that, for every positive integer k <= a(n), n^2 can be written as the sum of k positive square integers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 155, 1, 211, 1, 275, 1, 1, 2, 1, 1, 1, 1, 611, 662, 1, 1, 827, 886, 1, 1, 1, 1142, 1211, 1, 1355, 1, 1507, 2, 1667, 1, 1, 1, 2011, 1, 1, 1, 1, 2486, 2587, 2690, 2795, 1, 3011, 1, 1, 3350, 1, 3586, 3707, 1, 1, 1

Offset: 1

Bernard Schott, Aug 17 2019

nonn

The idea for this sequence comes from the 6th problem of the 2nd day of the 33rd International Mathematical Olympiad in Moscow, 1992 (see link).
There are four cases to examine and three possible values for a(n).
a(n) = 1 iff n is a nonhypotenuse number or iff n is in A004144.
a(n) >= 2 iff n is a hypotenuse number or iff n is in A009003.
a(n) = 2 iff n^2 is the sum of two positive squares but not the sum of three positive squares or iff n^2 is in A309779.
a(n) = n^2 - 14 iff n^2 is the sum of two and three positive squares or iff n^2 is in A231632.
Theorem: a square n^2 is the sum of k positive squares for all 1 <= k <= n^2 - 14 iff n^2 is the sum of 2 and 3 positive squares (proof in Kuczma). Consequently: A231632 = A018820.

			1 = 1^2, 4 = 2^2 and a(1) = a(2) = 1.
25 = 5^2 = 3^2 + 4^2 and a(5) = 2.
The first representations of 169 are 13^2 = 12^2 + 5^2 = 12^2 + 4^2 + 3^2 = 11^2 + 4^2 + 4^2 + 4^2 =  6^2 + 6^2 + 6^2 + 6^2 + 5^2  = 6^2 + 6^2 + 6^2 + 6^2 + 4^2 + 3^2 = ... and a(13) = 13^2 - 14 = 155.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 76-79.

IMO, 1992, Moscow, Second day. Problem 6

Cf. A018820, A004144, A009003, A231632, A309779.

cp's OEIS Frontend

A153894 a(n) = 5*2^n - 1.

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