Ron Knott - cp's OEIS frontend

A295089 a(n) = 3*n^2 + n + 3.

3, 7, 17, 33, 55, 83, 117, 157, 203, 255, 313, 377, 447, 523, 605, 693, 787, 887, 993, 1105, 1223, 1347, 1477, 1613, 1755, 1903, 2057, 2217, 2383, 2555, 2733, 2917, 3107, 3303, 3505, 3713, 3927, 4147, 4373, 4605, 4843, 5087, 5337, 5593, 5855, 6123, 6397, 6677, 6963, 7255, 7553, 7857

Offset: 0

Ron Knott, Nov 14 2017

Numbers represented as the palindrome 313 in number base n including base n=1, base 2 (binary) and base 3 with 'illegal' digit 3: 313_1=7, 313_2=17, 313_3=33, ... 313_9=255, 313_10=313, ...

			313 in base 7 is 3*7^2 + 1*7 + 3 = 157.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A049451, A056108, A131649, A144391.

Array[3 #^2 + # + 3 &, 52, 0] (* Michael De Vlieger, Nov 15 2017 *)
LinearRecurrence[{3, -3, 1}, {3, 7, 17}, 52] (* or *)
CoefficientList[Series[-(5 x^2 - 2 x + 3)/(x - 1)^3, {x, 0, 51}], x] (* Robert G. Wilson v, Nov 29 2017 *)

a(n) = 3*n^2 + n + 3; \\ Michel Marcus, Dec 15 2017

a(n) = A131649(n+3) + 1, n >= 2 (conjectured).
a(n) = A056108(n) + 2 = A049451(n) + 3 = A144391(n) + 4.
From Elmo R. Oliveira, Sep 02 2025: (Start)
G.f.: (3 - 2*x + 5*x^2)/(1-x)^3.
E.g.f.: (3 + 4*x + 3*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)

A277534 Least hypotenuse, c, of a Primitive Pythagorean Triangle (PPT) such that the difference between it, c, and its greater leg, b, is n; or 0 if no such PPT exists.

Original entry on oeis.org

5, 17, 0, 0, 65, 0, 0, 29, 65, 185, 0, 0, 169, 0, 0, 0, 221, 333, 0, 0, 273, 0, 0, 0, 157, 481, 0, 0, 1189, 0, 0, 641, 1353, 629, 0, 0, 1517, 0, 0, 425, 1681, 777, 0, 0, 1845, 0, 0, 0, 205, 925, 0, 0, 2173, 0, 0, 0, 2337, 1073, 0, 0, 2501, 0, 0, 0, 2665, 1221, 0, 0, 2829, 0, 0, 1405, 2993, 1369, 0

Offset: 1

Ron Knott and Robert G. Wilson v, Jun 05 2014

n = 1, 2, 5, 8, 9, 10, 13, 17, 18, 21, 25, ..., satisfies the first criterion;
a(n) = 0 for n = 3, 4, 6, 7, 11, 12, 14, 15, 16, 19, 20, 22, 23, 24, ..., ;
a(n) = 0 for 5832 of the first 10000 terms;
a(8n) = 0 for 832 of the first 10000 terms;
a(8n) = 0 for n: 2, 3, 6, 7, 8, 10, 11, 12, 14, 15, 18, 19, 22, 23, 24, ..., ;
a(8n+1) > 0;
a(8n+2) > 0; a linear 2nd-order recurrence: a(n) = 2*a(n-1) - a(n-2) with a(1) = 185 & a(2) = 333;
a(8n+3) = 0;
a(8n+4) = 0;
a(8n+5) > 0;
a(8n+6) = 0;
a(8n+7) = 0;
Prime terms: 5, 17, 29, 157, 641, 3821, 4201, 17749, 21601, 31981, 38273, 44789, 61129, 66173, 72161, 100673, 108541, 114553, 121421, 142973, 165541, 173777, 182141, 204733, 213881, 225889, 235493, 281837, ..., .

			a(1) is 5 since the PPT (3,4,5) satisfies the first stated criterion; a(2) is 17 since the PPT (8,15,17) satisfies the first stated criterion; a(3) = 0 since there exists no PPT that satisfies the stated criteria; etc.

Ron Knott and Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Ron Knott, Pythagorean Triples and Online Calculators

Cf. A008846, A020882, A242219.

f[n_] := FindInstance[ a^2 + b^2 == c^2 && Mod[c, 4] == 1 && 0 < a < b < c && c - b == n, {a, b, c}, Integers][[1, 3, 2, 1, 1, 3]] + 1 /. 1 + {}[[1, 3, 2, 1, 1, 3]] -> 0; f[1] = 5; Array[f, 75]

A226976 Fibonacci(n)^3 + Fibonacci(n+2)^3.

Original entry on oeis.org

1, 9, 28, 133, 539, 2322, 9773, 41501, 175636, 744273, 3152359, 13354306, 56568617, 239630337, 1015087436, 4299984173, 18215017507, 77160064914, 326855259829, 1384581132277, 5865179743556, 24845300179929, 105246380344463, 445830821750018, 1888569667033489

Offset: 0

Ron Knott, Jun 27 2013

nonn

			a(2) = Fibonacci(2)^3 + Fibonacci(4)^3 = 1^3 + 2^3 = 9

Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045 (Fibonacci), A056570 (Fibonacci^3).

Cf. A110224 (Fib(n)^3 + Fib(n+1)^3).

Table[Fibonacci[n]^3 + Fibonacci[n+2]^3,{n,0,50}]
#[[1]]+#[[3]]&/@Partition[Fibonacci[Range[0,30]]^3,3,1] (* or *) LinearRecurrence[{3,6,-3,-1},{1,9,28,133},30] (* Harvey P. Dale, Jun 30 2025 *)

a(n) = fibonacci(n)^3+fibonacci(n+2)^3; \\ Joerg Arndt, Jul 07 2013

a(n) = 3a(n-1)+6a(n-2)-3a(n-3)-a(n-4)

G.f.: (1+6x-5x^2-2x^3)/(1-3x-6x^2+3x^3+x^4)= (2x^2+7x+1)(1-x)/((x^2-x-1)(x^2+4x-1))

A226958 a(n) = Fibonacci(n-2)Fibonacci(n)Fibonacci(n+2).

Original entry on oeis.org

2, 0, 10, 24, 130, 504, 2210, 9240, 39338, 166320, 705058, 2985840, 12649570, 53582256, 226981610, 961503816, 4073004770, 17253510120, 73087065922, 309601740360, 1311494081482, 5555577978720, 23533806138050, 99690802301664, 422297015715650, 1788878864564064, 7577812474943050

Offset: 1

Ron Knott, Jun 27 2013

nonn

			a(3) = F(1)*F(3)*F(5) = 1*2*5 = 10.

Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers).

Products of 3 Fibonaccis: A065563, A056570, A220362, A110224.

Table[Fibonacci[n - 2] Fibonacci[n] Fibonacci[n + 2], {n, 1, 20}]
LinearRecurrence[{3,6,-3,-1},{2,0,10,24},30] (* Harvey P. Dale, Apr 10 2022 *)
Join[{2},#[[1]]#[[3]]#[[5]]&/@Partition[Fibonacci[Range[0,40]],5,1]] (* Harvey P. Dale, May 20 2025 *)

a(n)=fibonacci(n-2)*fibonacci(n)*fibonacci(n+2); \\ Joerg Arndt, Jul 07 2013

a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4).
G.f.: 2*(1-3*x-x^2)/(1-3*x-6*x^2+3*x^3+x^4).
a(n) = Lucas(n-1)*Fibonacci(n+2) = Fibonacci(n-2)*Lucas(n+1).
a(n) = (1/5)*(Fibonacci(3*n)-8*(-1)^n*Fibonacci(n)). - Ehren Metcalfe, Mar 26 2016
For n >= 3, a(n) is the numerator of the continued fraction [1,..,1, 3 ,1,..,1, 3 ,1,..,1] with three runs of 1's each of length n-3 and each separated by a single 3. For example, a(5)=130 which is the numerator of the continued fraction [1,1, 3 ,1,1, 3 ,1,1]. - Greg Dresden, Jan 01 2022

More terms from Joerg Arndt, Jul 07 2013

A147561 Number of representations of n in the Fibonacci-squared base system. The columns are ..., 64, 25, 9, 4, 1, 1 = ..., 8^2, 5^2, 3^2, 2^2, 1^2, 1^2, i.e., the Fibonacci numbers A000045 squared. The 'digits' are 0, 1 or 2.

Original entry on oeis.org

2, 3, 2, 2, 2, 3, 2, 2, 3, 5, 5, 3, 2, 2, 3, 2, 2, 3, 5, 5, 3, 2, 2, 3, 3, 4, 5, 5, 4, 3, 3, 2, 2, 3, 5, 5, 3, 2, 2, 3, 2, 2, 3, 5, 5, 3, 2, 2, 3, 3, 4, 5, 5, 4, 3, 3, 2, 2, 3, 5, 5, 3, 2, 3, 5, 5, 4, 5, 7, 8, 5, 4, 5, 8, 7, 5, 4, 5, 5, 3, 2, 3, 5, 5, 3, 2, 2, 3, 3, 4, 5, 5, 4, 3, 3, 2, 2, 3, 5, 5

Offset: 1

Ron Knott, Nov 07 2008

Note there are two columns labeled 1.

			a(2) = 3 since 2 is 02, 20 and 11 using both columns labeled 1;
a(10) = 5 because 10 = 9 + 1 with 2 Fib-sq reps 1010, 1001; 10 = 2*4 + 2 with 3 Fib-sq reps 220, 211 and 202; so there are in total 5 Fib-sq representations for 10.

David A. Corneth, Table of n, a(n) for n = 1..10000
Ron Knott, Fibonacci Bases: The Fibonacci^2 Base System

Cf. A000045, A007598.

first(n) = {my(fib2list = List(), fib2 = 1, t = 1, res = vector(n)); while(fib2 <= n, listput(fib2list, fib2); t++; fib2 = fibonacci(t)^2); for(i=1,3^#fib2list-1, b = digits(i,3); b = concat(vector(#fib2list-#b),b); s = sum(i=1,#b, b[i]*fib2list[i]); if(s<=n, res[s]++));res} \\ David A. Corneth, Jul 24 2017

A122194 Numbers that are the sum of exactly two sets of Fibonacci numbers.

Original entry on oeis.org

3, 5, 6, 9, 10, 15, 17, 25, 28, 41, 46, 67, 75, 109, 122, 177, 198, 287, 321, 465, 520, 753, 842, 1219, 1363, 1973, 2206, 3193, 3570, 5167, 5777, 8361, 9348, 13529, 15126, 21891, 24475, 35421, 39602, 57313, 64078, 92735, 103681, 150049, 167760, 242785

Offset: 1

Ron Knott, Aug 25 2006

			a(1)=3 as 3 is the sum of just 2 Fibonacci sets {3=Fibonacci(4)} and {1=Fibonacci(2), 2=Fibonacci(3)};
a(2)=5 as 5 is sum of Fibonacci sets {5} and {2,3} only.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
J. Berstel, An Exercise on Fibonacci Representations, RAIRO/Informatique Theorique, Vol. 35, No 6, 2001, pp. 491-498, in the issue dedicated to Aldo De Luca on the occasion of his 60th anniversary.
M. Bicknell-Johnson & D. C. Fielder, The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas, Fibonacci Quart. 37.1 (1999) pp. 47 ff.
Ron Knott Sumthing about Fibonacci Numbers
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1).

Cf. A000032, A000045, A000071, A000119, A013583, A122195.

a:= function(n)
    if n mod 2=0 then return 2*Fibonacci(Int((n+6)/2)) -1;
    else return Lucas(1,-1, Int((n+5)/2))[2] -1;
    fi;
  end;
List([1..50], n-> a(n) ); # G. C. Greubel, Jul 13 2019

f:=Floor; [(n mod 2) eq 0 select 2*Fibonacci(f((n+6)/2))-1 else Lucas(f((n+5)/2))-1: n in [1..50]]; // G. C. Greubel, Jul 13 2019

fib:= combinat[fibonacci]:
lucas:=n->fib(n-1)+fib(n+1):
a:=n -> if n mod 2 = 0 then 2 *fib(n/2+3) -1 else lucas((n+1)/2+2)-1 fi:
seq(a(n), n=1..50);

LinearRecurrence[{1, 1, -1, 1, -1}, {3, 5, 6, 9, 10, 15}, 40] (* Vincenzo Librandi, Jul 25 2017 *)
Table[If[Mod[n,2]==0, 2*Fibonacci[(n+6)/2]-1, LucasL[(n+5)/2]-1], {n,50}] (* G. C. Greubel, Jul 13 2019 *)

vector(50, n, f=fibonacci; if(n%2==0, 2*f((n+6)/2)-1, f((n+7)/2) + f((n+3)/2)-1)) \\ G. C. Greubel, Jul 13 2019

def a(n):
    if (mod(n,2)==0): return 2*fibonacci((n+6)/2) - 1
    else: return lucas_number2((n+5)/2, 1,-1) -1
[a(n) for n in (1..50)] # G. C. Greubel, Jul 13 2019

a(2n-1) = A000032(n+2) - 1,
a(2n) = 2*A000045(n+3) - 1.
a(2n-1) = A001610(n+2), a(2n) = A001595(n+2).
a(1)=3, a(2)=5, a(3)=6, a(4)=9, a(n) = a(n-2) + a(n-4) + 1, n > 4.
G.f.: (3 + 2*x - 2*x^2 + x^3 - 3*x^4)/(1-x-x^2+x^3-x^4+x^5).
a(n) = A272632(n)-1. - R. J. Mathar, Jan 13 2023

A122195 Numbers that are the sum of exactly 3 sets of Fibonacci numbers.

Original entry on oeis.org

8, 11, 13, 14, 18, 19, 22, 23, 30, 31, 36, 38, 49, 51, 59, 62, 80, 83, 96, 101, 130, 135, 156, 164, 211, 219, 253, 266, 342, 355, 410, 431, 554, 575, 664, 698, 897, 931, 1075, 1130, 1452, 1507, 1740, 1829, 2350, 2439, 2816, 2960, 3803, 3947, 4557, 4790, 6154

Offset: 1

Ron Knott, Aug 25 2006, corrected Aug 29 2006

nonn

			8 is the sum of only 3 sets of Fibonacci numbers: {8}, {3,5} and {1,2,5};
11 is the sum of only {3,8}, {1,2,8}, {1,2,3,5}.

G. C. Greubel, Table of n, a(n) for n = 1..1000
M. Bicknell-Johnson & D. C. Fielder, The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas, Fibonacci Quart. 37.1 (1999) pp. 47 ff.
Ron Knott, Sumthing about Fibonacci Numbers
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1,0,0,1,-1).

Cf. A000045, A000071, A013583, A000119, A122194.

a:=[11,13,14,18,19,22,23,30];; for n in [9..60] do a[n]:=a[n-4]+a[n-8]+1; od; Concatenation([8], a); # G. C. Greubel, Jul 13 2019

R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9) )); // G. C. Greubel, Jul 13 2019

# first N terms:
series((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(x^9-x^8+x^5-x^4-x+1),x,N+1);

CoefficientList[Series[(8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9), {x, 0, 60}], x] (* G. C. Greubel, Jul 13 2019 *)

my(x='x+O('x^60)); Vec((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8 -3*x^9)/(1-x-x^4+x^5-x^8+x^9)) \\ G. C. Greubel, Jul 13 2019

((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Jul 13 2019

G.f.: (8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1-x-x^4+x^5-x^8+x^9).
a(n) = a(n-4) + a(n-8) + 1.
a(0)=8, a(1)=11, a(2)=13, a(3)=18, then: a(4n) = A022318(n+3) = 2*A000045(n+5) + A000045(n+3) - 1, a(4n+1) = A022406(n+2) = 4*A000045(n+4) - 1, a(4n+2) = A022308(n+4) = 2*A000045(n+4) + A000045(n+6) - 1, a(4n+3) = 3*A000045(n+4) - 1, for n>=1.
a(n) = a(n-1) +a(n-4) -a(n-5) +a(n-8) -a(n-9). - G. C. Greubel, Jul 13 2019

A110703 Numbers S with two neighboring run sums (sum of positive integer runs) S = a+(a+1)+..+b=(b+1)+(b+2)...+c, 0

Original entry on oeis.org

3, 15, 27, 30, 42, 75, 90, 105, 135, 147, 165, 243, 252, 270, 273, 315, 363, 375, 378, 420, 462, 495, 507, 612, 660, 675, 693, 735, 750, 780, 810, 855, 858, 867, 945, 1050, 1083, 1155, 1170, 1215, 1287, 1323, 1365, 1470, 1485, 1518, 1587, 1785, 1815, 1875, 1950

Offset: 1

Ron Knott, Aug 04 2005

nonn

In other words, numbers n such that a list of consecutive numbers can be split into two parts in which their sums both equal n. - A. D. Skovgaard, May 22 2017
If the two runs overlap in one number, the runs are Friends and their sums are A110701. The sums are the difference of two triangular numbers A000217.
The subsequence where there is more than one possible splitting begins 105, 945, 1365, 2457, 2625, 3990, 5145, 8505, ... - Jean-François Alcover, May 22 2017
a(n) seems to always be divisible by 3.- A. D. Skovgaard, May 22 2017. This is true. Sequence lists values of n = t(t+1)/2 - k(k+1)/2 = m(m+1)/2 - t(t+1)/2 with k < t < m. Since any triangular number must be of the form 3w or 3w+1, then there are two possibilities for n = 3w - k(k+1)/2 = m(m+1)/2 - 3w or n = 3w + 1 - k(k+1)/2 = m(m+1)/2 - 3w - 1. For first case, if k(k+1)/2 = 3u+1, there is no solution for m. Similarly for second case, if k(k+1)/2 = 3u, there is no solution for m. So always n must be divisible by 3. - Altug Alkan, May 22 2017

			3 = 1+2 = 3, so 3 is a term.
15 = 4+5+6 = 7+8 so 15 is a term.
a(6) = 75 because 75 = 3+4+5+6+7+8+9+10+11+12 = 13+14+15+16+17.

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110701, A110702.

Select[Range[1000], False =!= Reduce[# == Sum[k, {k, x, y}] == Sum[k, {k, y + 1, z}] && z >= y >= x > 0, {x, y, z}, Integers] &] (* Giovanni Resta, May 22 2017 *)

Initial 3 added by A. D. Skovgaard, May 22 2017

A110701 Friendly run sums: numbers S with two run sums (sum of positive integer runs) that share one common number, i.e., S = a + (a+1) + ... + b = b + (b+1) + ... + c with a < b < c.

Original entry on oeis.org

9, 21, 30, 42, 65, 70, 99, 105, 117, 133, 135, 154, 175, 180, 225, 231, 275, 285, 341, 342, 345, 364, 385, 414, 440, 450, 455, 481, 495, 540, 546, 567, 630, 645, 675, 693, 744, 750, 765, 825, 833, 936, 945, 990, 1035, 1045, 1140, 1161, 1170, 1176, 1178

Offset: 1

Ron Knott, Aug 04 2005

nonn

The sums are the difference of two triangular numbers A000217. The common numbers themselves are A094550. The sums, n > 0, are n = (b-a+1)(a+b)/2 = (b+c)(c-b+1)/2 where b^2 + a - a^2 = c^2 + c - b^2, is solvable in integers for 0 < a < b < c. Since the runs have something in common, they are "friends". The series of sums *without* the common number is A110702. The numbers common to the two runs are A094550.

			2 + 3 + 4 = 4 + 5 share the common number 4. The sum of both is 9 and this is the smallest such sum with a common "friend" (4), so a(1)=9.

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110702, A110703.

A110702 Numbers S with two runsums (sum of positive integer runs) where the 2 runs are separated by a single number gap (b), i.e., S = a + (a+1) + ... + (b-1) = (b+1) + ... + c with a < b-1, b < c.

Original entry on oeis.org

5, 15, 21, 33, 54, 54, 85, 90, 100, 117, 110, 135, 153, 161, 204, 195, 252, 261, 315, 308, 315, 315, 351, 385, 405, 418, 414, 450, 455, 476, 510, 533, 595, 609, 637, 650, 705, 693, 684, 777, 792, 870, 884, 931, 986, 999, 1071, 1105, 1121, 1110, 1125, 1210, 1230

Offset: 1

Ron Knott, Aug 04 2005

nonn

The sums are the difference of two triangular numbers A000217. The sum series where the missing number is included is A110701. The numbers in the gap between the two runs are A094550.

			1+2+3+4+5 and 7+8 have a gap between them of (6). The sum of both is 15 so 15 is in the ordered series (as a(2)).

Ron Knott Runsums
T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

Cf. A001227, A094550, A110701, A110703.

cp's OEIS Frontend

User: Ron Knott

A295089 a(n) = 3*n^2 + n + 3.

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A277534 Least hypotenuse, c, of a Primitive Pythagorean Triangle (PPT) such that the difference between it, c, and its greater leg, b, is n; or 0 if no such PPT exists.

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A226976 Fibonacci(n)^3 + Fibonacci(n+2)^3.

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A226958 a(n) = Fibonacci(n-2)*Fibonacci(n)*Fibonacci(n+2).

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A147561 Number of representations of n in the Fibonacci-squared base system. The columns are ..., 64, 25, 9, 4, 1, 1 = ..., 8^2, 5^2, 3^2, 2^2, 1^2, 1^2, i.e., the Fibonacci numbers A000045 squared. The 'digits' are 0, 1 or 2.

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A122194 Numbers that are the sum of exactly two sets of Fibonacci numbers.

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A122195 Numbers that are the sum of exactly 3 sets of Fibonacci numbers.

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A226958 a(n) = Fibonacci(n-2)Fibonacci(n)Fibonacci(n+2).