A138986 -id:A138986 - cp's OEIS frontend

A138997 First differences of Frobenius numbers for 6 successive numbers A138986.

1, 1, 1, 1, 8, 2, 2, 2, 2, 14, 3, 3, 3, 3, 20, 4, 4, 4, 4, 26, 5, 5, 5, 5, 32, 6, 6, 6, 6, 38, 7, 7, 7, 7, 44, 8, 8, 8, 8, 50, 9, 9, 9, 9, 56, 10, 10, 10, 10, 62, 11, 11, 11, 11, 68, 12, 12, 12, 12, 74, 13, 13, 13, 13, 80, 14, 14, 14, 14, 86, 15, 15, 15, 15, 92, 16, 16, 16, 16, 98, 17, 17

Offset: 1

Artur Jasinski, Apr 05 2008

nonn

For first differences of Frobenius numbers for 2 successive numbers see A005843
For first differences of Frobenius numbers for 3 successive numbers see A014682
For first differences of Frobenius numbers for 4 successive numbers see A138995
For first differences of Frobenius numbers for 5 successive numbers see A138996
For first differences of Frobenius numbers for 6 successive numbers see A138997
For first differences of Frobenius numbers for 7 successive numbers see A138998
For first differences of Frobenius numbers for 8 successive numbers see A138999

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).

Cf. A028387, A037165, A079326, A138985, A138986, A138987, A138988, A138989, A138990, A138991, A138992, A138993, A138994, A138995, A138996, A138997, A138998, A138999.

a = {}; Do[AppendTo[a, FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4, n + 5, n + 6}]], {n, 1, 100}]; Differences[a]
LinearRecurrence[{0, 0, 0, 0, 2, 0, 0, 0, 0, -1}, {1, 1, 1, 1, 8, 2,
  2, 2, 2, 14}, 50] (* G. C. Greubel, Feb 18 2017 *)
Differences[Table[FrobeniusNumber[Range[n,n+5]],{n,2,90}]] (* Harvey P. Dale, Dec 18 2023 *)

x='x + O('x^50); Vec(-(-1-x-x^2-x^3-8*x^4+2*x^9)/((x-1)^2*(x^4+x^3+x^2+x+1)^2)) \\ G. C. Greubel, Feb 18 2017

a(n) = A138986(n+1) - A138986(n).
O.g.f.= -(-1-x-x^2-x^3-8*x^4+2*x^9)/((x-1)^2*(x^4+x^3+x^2+x+1)^2). - R. J. Mathar, Apr 20 2008
a(n) = 2*a(n-5) - a(n-10). - R. J. Mathar, Apr 20 2008
a(n)= (1/5)*n*x(5+mod(n,5))-(1/5)*mod(n,5)*x(5+mod(n,5))+x(mod(n,5))-(1/5)*n*x(mod(n,5))+(1/5) *mod(n,5)*x(mod(n,5)). - Alexander R. Povolotsky, Apr 20 2008

A028387 a(n) = n + (n+1)^2.

Original entry on oeis.org

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651

Offset: 0

Patrick De Geest

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, World!Of Numbers, Palindromes of the form n+(n+1)^2.
Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Nandini Nilakantan and Anurag Singh, Homotopy type of neighborhood complexes of Kneser graphs, KG_{2,k}, Proceeding-Mathematical Sciences, 128, Article number: 53(2018).
Yanni Pei and Jiang Zeng, Counting signed derangements with right-to-left minima and excedances, arXiv:2206.11236 [math.CO], 2022.
Popular Computing (Calabasas, CA), The CSR Function, Vol. 4 (No. 34, Jan 1976), pages PC34-10 to PC34-11. Annotated and scanned copy.
Zdzislaw Skupień and Andrzej Żak, Pair-sums packing and rainbow cliques, in Topics In Graph Theory, A tribute to A. A. and T. E. Zykovs on the occasion of A. A. Zykov's 90th birthday, ed. R. Tyshkevich, Univ. Illinois, 2013, pages 131-144, (in English and Russian).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
Cf. also A135223, A176271, A214604, A105728, A005408, A002378, A084990, A253607, A260910, A089270.

a028387 n = n + (n + 1) ^ 2  -- Reinhard Zumkeller, Jul 17 2014

[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *)
Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)

a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024

[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A138985 a(n) = Frobenius number for 5 successive numbers = F(n+1, n+2, n+3, n+4, n+5).

Original entry on oeis.org

1, 2, 3, 4, 11, 13, 15, 17, 29, 32, 35, 38, 55, 59, 63, 67, 89, 94, 99, 104, 131, 137, 143, 149, 181, 188, 195, 202, 239, 247, 255, 263, 305, 314, 323, 332, 379, 389, 399, 409, 461, 472, 483, 494, 551, 563, 575, 587, 649, 662, 675, 688, 755, 769, 783, 797, 869

Offset: 1

Artur Jasinski, Apr 05 2008

			a(5)=11 because 11 is the largest number k such that the equation 6*x_1 + 7*x_2 + 8*x_3 + 9*x_4 + 10*x_5 = k has no solution for any nonnegative x_i (in other words, for every k > 11 there exist one or more solutions).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,2,-2,0,0,-1,1).

Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), this sequence (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Table[FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4}], {n, 1, 100}]
Table[(Floor[(n-1)/4]+1)*(n+1)-1,{n,57}] (* Zak Seidov, Jan 10 2015 *)
FrobeniusNumber/@Partition[Range[2,70],5,1] (* or *) LinearRecurrence[ {1,0,0,2,-2,0,0,-1,1},{1,2,3,4,11,13,15,17,29},70] (* Harvey P. Dale, Oct 07 2016 *)

for (n=1,57,print1((floor((n-1)/4)+1)*(n+1)-1 ","))\\ Zak Seidov, Jan 10 2015

G.f.: x*(x^8 - 5*x^4 - x^3 - x^2 - x - 1) / ((x-1)^3*(x+1)^2*(x^2+1)^2). - Colin Barker, Dec 13 2012

A079326 a(n) = the largest number m such that if m monominoes are removed from an n X n square then an L-tromino must remain.

Original entry on oeis.org

1, 2, 7, 9, 17, 20, 31, 35, 49, 54, 71, 77, 97, 104, 127, 135, 161, 170, 199, 209, 241, 252, 287, 299, 337, 350, 391, 405, 449, 464, 511, 527, 577, 594, 647, 665, 721, 740, 799, 819, 881, 902, 967, 989, 1057, 1080, 1151, 1175, 1249, 1274, 1351, 1377, 1457

Offset: 2

Mambetov Timur (timur_teufel(AT)mail.ru), Feb 13 2003

			a(3)=2 because if a middle row of 3 monominoes are removed from the 3 X 3, no L remains.

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Frobenius number for k successive numbers: A028387 (k=2), this sequence (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Cf. A093353, A104519.

Table[FrobeniusNumber[{a, a + 1, a + 2}], {a, 2, 54}] (* Zak Seidov, Jan 08 2015 *)

a(n) = (n^2)/2 - 1 (n even), (n^2-n)/2 - 1 (n odd).
a(n) = A204557(n-1) / (n-1). - Reinhard Zumkeller, Jan 18 2012
From Bruno Berselli, Jan 18 2011: (Start)
G.f.: x^2*(1+x+3*x^2-x^4)/((1+x)^2*(1-x)^3).
a(n) = n*(2*n+(-1)^n-1)/4 - 1.
a(n) = A105638(-n+2). (End)

Edited by Don Reble, May 28 2007

A138987 a(n) = Frobenius number for 7 successive numbers = F(n+1, n+2, n+3, n+4, n+5, n+6, n+7).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 15, 17, 19, 21, 23, 25, 41, 44, 47, 50, 53, 56, 79, 83, 87, 91, 95, 99, 129, 134, 139, 144, 149, 154, 191, 197, 203, 209, 215, 221, 265, 272, 279, 286, 293, 300, 351, 359, 367, 375, 383, 391, 449, 458, 467, 476, 485, 494, 559, 569, 579, 589, 599

Offset: 1

Artur Jasinski, Apr 05 2008

			a(7) = 15 because 15 is the largest number k such that the equation 8*x_1 + 9*x_2 + 10*x_3 + 11*x_4 + 12*x_5 + 13*x_6 + 14*x_7 = k has no solution for any nonnegative x_i (in other words, for every k > 15 there exist one or more solutions).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,2,-2,0,0,0,0,-1,1).

Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), this sequence (k=7), A138988 (k=8).

Table[FrobeniusNumber[{n+1, n+2, n+3, n+4, n+5, n+6, n+7}], {n, 1, 100}]
Table[FrobeniusNumber[n+Range[7]],{n,100}] (* Harvey P. Dale, Dec 06 2021 *)
Table[n + Floor[(n-1)/6]*(n+1), {n, 100}] (* Giorgos Kalogeropoulos, Apr 06 2025 *)

G.f.: x*(x^12-7*x^6-x^5-x^4-x^3-x^2-x-1) / ((x-1)^3*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2). [Colin Barker, Dec 13 2012]

a(n) = n + (n+1)*floor((n-1)/6). - Giorgos Kalogeropoulos, Apr 06 2025

A138988 a(n) is the Frobenius number for 8 successive numbers n+1, n+2, ..., n+8.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 17, 19, 21, 23, 25, 27, 29, 47, 50, 53, 56, 59, 62, 65, 91, 95, 99, 103, 107, 111, 115, 149, 154, 159, 164, 169, 174, 179, 221, 227, 233, 239, 245, 251, 257, 307, 314, 321, 328, 335, 342, 349, 407, 415, 423, 431, 439, 447, 455, 521, 530, 539

Offset: 1

Artur Jasinski, Apr 05 2008

			a(8)=17 because 17 is the largest number k such that equation:
9*x_1 + 10*x_2 + 11*x_3 + 12*x_4 + 13*x_5 + 14*x_6 + 15*x_7 + 16*x_8 = k has no solution for any nonnegative x_i (in other words, for every k > 17 there exist one or more solutions).

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,2,-2,0,0,0,0,0,-1,1).

Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), this sequence (k=8).

Table[FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4, n + 5, n + 6, n + 7, n + 8}], {n, 1, 100}]
Table[FrobeniusNumber[n+Range[8]],{n,100}] (* Harvey P. Dale, Sep 22 2015 *)

G.f.: x*(x^14 - 8*x^7 - x^6 - x^5 - x^4 - x^3 - x^2 - x - 1) / ((x-1)^3*(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)^2). - Colin Barker, Dec 13 2012

A138990 a(n) = Frobenius number for 4 successive primes = F[p(n), p(n+1), p(n+2), p(n+3)].

Original entry on oeis.org

1, 4, 9, 23, 42, 67, 83, 101, 125, 199, 262, 335, 367, 393, 492, 593, 704, 807, 873, 990, 817, 950, 1101, 1353, 2039, 2624, 2371, 1494, 1431, 1640, 2927, 2368, 2875, 2667, 3570, 3348, 3625, 3918, 4531, 3816, 4831, 4543, 9357, 4819, 4131, 6611, 5735, 10483

Offset: 1

Artur Jasinski, Apr 05 2008

nonn

			a(3)=23 because 23 is the largest number k such that the equation 7*x_1 + 11*x_2 + 13*x_3 + 17*x + 4 = k has no solution for any nonnegative x_i (in other words, for every k > 23 there exist one or more solutions).

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Frobenius numbers for k successive primes: A037165 (k=2), A138989 (k=3), this sequence (k=4), A138991 (k=5), A138992 (k=6), A138993 (k=7), A138994 (k=8).

Cf. A028387, A079326, A138985, A138986, A138987, A138988.

Table[FrobeniusNumber[{Prime[n],Prime[n + 1], Prime[n + 2], Prime[n + 3]}], {n, 1, 100}]
FrobeniusNumber/@Partition[Prime[Range[60]],4,1] (* Harvey P. Dale, Nov 23 2014 *)

Definition corrected by Harvey P. Dale, Aug 15 2014

A138989 a(n) = Frobenius number for 3 successive primes = F[p(n), p(n+1), p(n+2)].

Original entry on oeis.org

1, 4, 13, 30, 53, 80, 117, 131, 194, 286, 293, 520, 613, 522, 1310, 858, 1001, 929, 1610, 1418, 1322, 1499, 1421, 2941, 3300, 3533, 3710, 3957, 2065, 2241, 3685, 4595, 3697, 3930, 5956, 12074, 5509, 5874, 14690, 7968, 6084, 6373, 12413, 12740, 6694, 21878

Offset: 1

Artur Jasinski, Apr 05 2008

nonn

			a(3)=13 because 13 is the largest number k such that the equation 5*x_1 + 7*x_2 + 11*x_3 = k has no solution for any nonnegative x_i. (In other words, for every k > 13 there exist one or more solutions.)

Frobenius numbers for k successive primes: A037165 (k=2), this sequence (k=3), A138990 (k=4), A138991 (k=5), A138992 (k=6), A138993 (k=7), A138994 (k=8).

Cf. A028387, A079326, A138985, A138986, A138987, A138988.

Table[FrobeniusNumber[{Prime[n],Prime[n + 1], Prime[n + 2]}], {n, 1, 100}]
FrobeniusNumber/@Partition[Prime[Range[50]],3,1] (* Harvey P. Dale, Dec 01 2015 *)

A138991 a(n) = Frobenius number for 5 successive primes = F[p(n), p(n+1), p(n+2), p(n+3), p(n+4)].

Original entry on oeis.org

1, 4, 9, 23, 31, 54, 66, 101, 125, 143, 200, 261, 285, 307, 398, 434, 588, 563, 672, 708, 659, 717, 935, 1078, 1748, 1816, 1135, 1173, 1104, 1277, 1911, 1975, 2188, 2111, 2680, 2593, 2683, 3266, 2861, 3297, 3757, 3996, 4198, 3275, 2953, 3457, 4668, 6688

Offset: 1

Artur Jasinski, Apr 05 2008

nonn

			a(3)=23 because 23 is the largest number k such that the equation 7*x_1 + 11*x_2 + 13*x_3 + 17*x_4 + 19*x_5 = k has no solution for any nonnegative x_i (in other words, for every k > 23 there exist one or more solutions).

Frobenius numbers for k successive primes: A037165 (k=2), A138989 (k=3), A138990 (k=4), this sequence (k=5), A138992 (k=6), A138993 (k=7), A138994 (k=8).

Cf. A028387, A079326, A138985, A138986, A138987, A138988.

Table[FrobeniusNumber[{Prime[n],Prime[n + 1], Prime[n + 2], Prime[n + 3], Prime[n + 4]}], {n, 1, 100}]
FrobeniusNumber/@Partition[Prime[Range[80]],5,1] (* Harvey P. Dale, Aug 15 2014 *)

A138992 a(n) = Frobenius number for 6 successive primes = F[p(n), p(n+1), p(n+2), p(n+3), p(n+4), p(n+5)].

Original entry on oeis.org

1, 4, 9, 16, 31, 41, 64, 63, 102, 143, 169, 216, 203, 264, 304, 381, 470, 502, 538, 562, 592, 638, 769, 989, 1360, 1008, 929, 961, 995, 1051, 1530, 1582, 1777, 1694, 2084, 2140, 2369, 2288, 2527, 2778, 3399, 2721, 2859, 2698, 2756, 3035, 3613, 5800, 4765

Offset: 1

Artur Jasinski, Apr 05 2008

nonn

			a(4)=16 because 16 is the largest number k such that equation 7*x_1 + 11*x_2 + 13*x_3 + 17*x_4 + 19*x_5 + 23*x_6 = k has no solution for any nonnegative x_i (in other words, for every k > 16 there exist one or more solutions).

Frobenius numbers for k successive primes: A037165 (k=2), A138989 (k=3), A138990 (k=4), A138991 (k=5), this sequence (k=6), A138993 (k=7), A138994 (k=8).

Cf. A028387, A079326, A138985, A138986, A138987, A138988.

Table[FrobeniusNumber[{Prime[n],Prime[n + 1], Prime[n + 2], Prime[n + 3], Prime[n + 4], Prime[n + 5]}], {n, 1, 100}]
FrobeniusNumber/@Partition[Prime[Range[100]],6,1] (* Harvey P. Dale, Aug 15 2014 *)

cp's OEIS Frontend

A138997 First differences of Frobenius numbers for 6 successive numbers A138986.

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A028387 a(n) = n + (n+1)^2.

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A138985 a(n) = Frobenius number for 5 successive numbers = F(n+1, n+2, n+3, n+4, n+5).

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A079326 a(n) = the largest number m such that if m monominoes are removed from an n X n square then an L-tromino must remain.

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A138987 a(n) = Frobenius number for 7 successive numbers = F(n+1, n+2, n+3, n+4, n+5, n+6, n+7).

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A138988 a(n) is the Frobenius number for 8 successive numbers n+1, n+2, ..., n+8.

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A138990 a(n) = Frobenius number for 4 successive primes = F[p(n), p(n+1), p(n+2), p(n+3)].

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