A278310
Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.
Original entry on oeis.org
3, 143, 4899, 166463, 5654883, 192099599, 6525731523, 221682772223, 7530688524099, 255821727047183, 8690408031080163, 295218051329678399, 10028723337177985443, 340681375412721826703, 11573138040695364122499, 393146012008229658338303, 13355391270239113019379843
Offset: 1
3 is in the sequence because T(3) + 3*T(4) = 6 + 3*10 = 6^2.
For n=5 is a(5) = 5654883, therefore floor(sqrt(5654883)) = 2377 = A182189(5) - 2 = 2379 - 2.
Cf.
A278438: numbers m such that T(m) + 2*T(m+1) is a square.
Cf.
A078522: numbers m such that 3*T(m) + T(m+1) is a square.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4:
A084703 (k=-1),
A076218 (k=3), this sequence (k=-5).
-
Iv:=[3,143]; [n le 2 select Iv[n] else 34*Self(n-1)-Self(n-2)+40: n in [1..20]];
-
P:=proc(q) local n; for n from 3 to q do if type(sqrt(2*n^2+5*n+3),integer) then print(n); fi; od; end: P(10^9); # Paolo P. Lava, Nov 18 2016
-
Table[((1 + Sqrt[2])^(4 n) + (1 - Sqrt[2])^(4 n))/8 - 5/4, {n, 1, 20}]
RecurrenceTable[{a[1] == 3, a[2] == 143, a[n] == 34 a[n - 1] - a[n - 2] + 40}, a, {n, 1, 20}]
LinearRecurrence[{35, -35, 1}, {3, 143, 4899}, 50] (* G. C. Greubel, Nov 20 2016 *)
-
Vec(x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)) + O(x^50)) \\ G. C. Greubel, Nov 20 2016
-
def A278310():
a, b = 3, 143
yield a
while True:
yield b
a, b = b, 34*b - a + 40
a = A278310(); print([next(a) for in range(18)]) # _Peter Luschny, Nov 18 2016
A285955
Numbers a(n) = T(b(n))*sqrt(T(b(n))+1), where T(b(n)) is the triangular number of b(n)= A000217(b(n)) and b(n)=A006451(n). Also a(n) = y solutions of the Bachet Mordell equation y^2=x^3+K, where x= T(b(n)) = A006454(n) and K = (T(b(n)))^2= A285985(n).
Original entry on oeis.org
0, 6, 60, 1320, 12144, 262080, 2405970, 51894744, 476378760, 10274921850, 94320640056, 2034382775040, 18675010652760, 402797515372356, 3697557790357470, 79751873665825680, 732097767490332144, 15790468188346521390, 144951660405354891060, 3126432949419110989944
Offset: 0
For n=2, b(n)=5, a(n)=60.
For n=5, b(n)=90, a(n)= 262080.
For n = 3, A006451(n) = 15. Therefore, A000217(A006451(n)) = A000217(15) = 120. This gives A000217(A006451(n)) * sqrt(A000217(A006451(n)) + 1) = 120 * sqrt(120 + 1) = 1320. - _David A. Corneth_, Apr 29 2017
- V. Pletser, On some solutions of the Bachet-Mordell equation for large parameter values, to be submitted, April 2017.
-
restart: bm2:=-1: bm1:=0: bp1:=2: bp2:=5: print (‘0,0’,’1,6’,’2,60’); for n from 3 to 1000 do b:= 8*sqrt((bp1^2+bp1)/2+1)+bm2; a:=(b*(b+1)/2)* sqrt((b*(b+1)/2)+1); print(n,a); bm2:=bm1; bm1:=bp1; bp1:=bp2; bp2:=b; end do:
A309507
Number of ways the n-th triangular number T(n) = A000217(n) can be written as the difference of two positive triangular numbers.
Original entry on oeis.org
0, 1, 1, 1, 3, 3, 1, 2, 5, 3, 3, 3, 3, 7, 3, 1, 5, 5, 3, 7, 7, 3, 3, 5, 5, 7, 7, 3, 7, 7, 1, 3, 7, 7, 11, 5, 3, 7, 7, 3, 7, 7, 3, 11, 11, 3, 3, 5, 8, 11, 7, 3, 7, 15, 7, 7, 7, 3, 7, 7, 3, 11, 5, 3, 15, 7, 3, 7, 15, 7, 5, 5, 3, 11, 11, 7, 15, 7, 3, 9, 9, 3, 7
Offset: 1
a(5) = 3: T(5) = T(6)-T(3) = T(8)-T(6) = T(15)-T(14).
a(7) = 1: T(7) = T(28)-T(27).
a(8) = 2: T(8) = T(13)-T(10) = T(36)-T(35).
a(9) = 5: T(9) = T(10)-T(4) = T(11)-T(6) = T(16)-T(13) = T(23)-T(21) = T(45)-T(44).
a(49) = 8: T(49) = T(52)-T(17) = T(61)-T(36) = T(94)-T(80) = T(127)-T(117) = T(178)-T(171) = T(247)-T(242) = T(613)-T(611) = T(1225)-T(1224).
The triples with n <= 16 are:
2, 2, 3
3, 5, 6
4, 9, 10
5, 3, 6
5, 6, 8
5, 14, 15
6, 5, 8
6, 9, 11
6, 20, 21
7, 27, 28
8, 10, 13
8, 35, 36
9, 4, 10
9, 6, 11
9, 13, 16
9, 21, 23
9, 44, 45
10, 8, 13
10, 26, 28
10, 54, 55
11, 14, 18
11, 20, 23
11, 65, 66
12, 17, 21
12, 24, 27
12, 77, 78
13, 9, 16
13, 44, 46
13, 90, 91
14, 5, 15
14, 11, 18
14, 14, 20
14, 18, 23
14, 33, 36
14, 51, 53
14, 104, 105
15, 21, 26
15, 38, 41
15, 119, 120
16, 135, 136. - _N. J. A. Sloane_, Mar 31 2020
- Alois P. Heinz, Table of n, a(n) for n = 1..20000
- J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
- M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.
See also
A053141. The monotonic triples [n,k,m] with n <= k <= m are counted in
A333529.
-
with(numtheory): seq(tau(n*(n+1))-tau(n*(n+1)/2)-1, n=1..80); # Ridouane Oudra, Dec 08 2023
-
TriTriples[TNn_] := Sort[Select[{TNn, (TNn + TNn^2 - # - #^2)/(2 #),
(TNn + TNn^2 - # + #^2)/(2 #)} & /@
Complement[Divisors[TNn (TNn + 1)], {TNn}],
And[And @@ (IntegerQ /@ #), And @@ (# > 0 & /@ #)] &]]
Length[TriTriples[#]] & /@ Range[100]
(* Bradley Klee, Mar 01 2020 *)
A373330
a(n) is the difference between T = A000217(n^2) and the greatest square not exceeding T.
Original entry on oeis.org
0, 0, 1, 9, 15, 1, 41, 0, 55, 72, 9, 156, 36, 204, 262, 144, 135, 289, 209, 316, 111, 117, 406, 309, 527, 261, 342, 860, 804, 36, 954, 1200, 624, 605, 1257, 969, 1400, 741, 849, 1856, 1639, 0, 1721, 2076, 855, 701, 1770, 1101, 1719, 397, 426, 1980, 1416, 2449, 1142
Offset: 0
A373331 and
A373332 are the coordinates of the observed lower envelope of this sequence.
-
Array[PolygonalNumber[#^2] - Floor[Sqrt[(#^4 + #^2)/2]]^2 &, 55, 0] (* Michael De Vlieger, Jun 02 2024 *)
-
a(n) = my(T=(n^4+n^2)/2); T-sqrtint(T)^2
-
from sympy import integer_nthroot
def A373330(n): return (T:=(n**4 + n**2) // 2)-(integer_nthroot(T,2)[0])**2
# Karl-Heinz Hofmann, Jul 01 2024
A025555
Least common multiple (or LCM) of first n positive triangular numbers (A000217).
Original entry on oeis.org
1, 3, 6, 30, 30, 210, 420, 1260, 1260, 13860, 13860, 180180, 180180, 180180, 360360, 6126120, 6126120, 116396280, 116396280, 116396280, 116396280, 2677114440, 2677114440, 13385572200, 13385572200, 40156716600, 40156716600
Offset: 1
a(5) = lcm{1, 3, 6, 10, 15} = 30.
-
a025555 n = a025555_list !! (n-1)
a025555_list = scanl1 lcm $ tail a000217_list
-- Reinhard Zumkeller, Nov 22 2013
-
HalfFarey := proc (n) local a,b,c,d,k,s; if n<2 then RETURN([1]) fi; a:=0; b:=1; c:=1; d:=n; s:=NULL; do k := iquo(n+b,d); a,b,c,d := c, d, k*c-a, k*d-b; if b < 2*a then break fi; s := s, a/b od; [s] end:
A025555 := proc(n) local r; HalfFarey(n+1); subsop(nops(%) = NULL,%); mul(2*sin(Pi*r),r = %)^2 end: seq(round(evalf(A025555(i))),i=1..27); # Peter Luschny, Jun 09 2011
-
nn=30;With[{trnos=Accumulate[Range[nn]]},Table[LCM@@Take[trnos,n], {n,nn}]] (* Harvey P. Dale, Oct 21 2011 *)
f[x_] := x + 1; a[1] = f[1]; a[n_] := LCM[f[n], a[n - 1]]; Array[a, 30]/2 (* Robert G. Wilson v, Jan 04 2013 *)
-
S=1;for(n=1,20,S=lcm(S,n*(n+1)/2);print1(S,",")) \\ Edward Jiang, Sep 08 2014
Original entry on oeis.org
0, 0, 0, 1, -1, 0, 4, 7, -7, -6, 0, 3, 13, 18, 28, 35, -35, -34, -24, -21, -5, 0, 14, 21, 43, 52, 70, 81, 105, 118, 140, 155, -155, -154, -136, -133, -105, -100, -78, -71, -35, -26, 0, 11, 47, 60, 90, 105, 151, 168, 202, 221, 265, 286, 324, 347, 399, 424, 466, 493, 545, 574, 620, 651, -651, -650, -616, -613, -561
Offset: 0
-
A048702 := Join[{0}, Reap[For[k = 1, k < 1500, k += 2, bb = IntegerDigits[k, 2]; If[bb == Reverse[bb], If[EvenQ[Length[bb]], Sow[k/3]]]]][[2, 1]]]; Table[n*(n + 1)/2 - A048702[[n + 1]], {n, 0, 50}] (* G. C. Greubel, Sep 26 2017 *)
-
a01(n) = my(f); f = length(binary(n)) - 1; 2^(f+1)*n + sum(i=0, f, bittest(n, i) * 2^(f-i)); \\ A048701
a(n) = n*(n+1)/2 - a01(n)/3; \\ A006095
-
def A075113(n: int) -> int:
s = bin(n)[2:]
return n * (n + 1) // 2 - int(s + s[::-1], 2) // 3
print([A075113(n) for n in range(69)]) # Peter Luschny, Dec 14 2022
Original entry on oeis.org
1, 1, 3, 6, 15, 28, 66, 120, 253, 465, 903, 1596, 3003, 5151, 9180, 15576, 26796, 44253, 74305, 120295, 196878, 314028, 502503, 788140, 1241100, 1917861, 2968266, 4531555, 6913621, 10421895, 15705210, 23409903, 34857075, 51445296, 75774205, 110759286
Offset: 0
-
f:= proc(n) local p;
p:= combinat:-numbpart(n);
p*(p+1)/2
end proc:
map(f, [$1..100]); # Robert Israel, Oct 26 2015
-
pp = Array[PartitionsP, 40, 0]; pp (pp + 1)/2 (* Jean-François Alcover, Mar 19 2019 *)
-
a(n) = apply(x->x*(x+1)/2, numbpart(n)); \\ Michel Marcus, Oct 26 2015
A092858
"Sum" of the sequences of primes and the triangular numbers (A000217).
Original entry on oeis.org
5, 6, 7, 10, 11, 13, 15, 17, 19, 21, 23, 28, 29, 31, 36, 37, 41, 43, 45, 47, 53, 55, 59, 61, 66, 67, 71, 73, 78, 79, 83, 89, 91, 97, 101, 103, 105, 107, 109, 113, 120, 127, 131, 136, 137, 139, 149, 151, 153, 157, 163, 167, 171, 173, 179, 181, 190, 191, 193, 197, 199
Offset: 1
Ferenc Adorjan (fadorjan(AT)freemail.hu)
A092859
"Difference" of the sequences of triangular numbers (A000217) and the primes (cf. A092858).
Original entry on oeis.org
3, 4, 5, 7, 12, 13, 16, 18, 19, 22, 23, 30, 31, 38, 39, 40, 42, 43, 46, 48, 49, 50, 51, 52, 53, 56, 57, 58, 60, 61, 68, 69, 70, 72, 73, 80, 81, 82, 84, 85, 86, 87, 88, 89, 92, 93, 94, 95, 96, 98, 99, 100, 102, 103, 106, 108, 110, 111, 112, 113, 121, 122, 123, 124, 125, 126
Offset: 1
Ferenc Adorjan (fadorjan(AT)freemail.hu)
-
{sdif(a,b)= /*Returns the "difference" of monotonic sequences a and b */ return(mtinv(mt(a)+mt(compl(b)))) /* the functions mt(a) and mtinv(r) are defined in A051006 and A092855, respectively */ } {compl(v)=/* Returns the complement of v monotonic positive sequence */ local(n,p=0,vv=[]);n=matsize(v)[2];for(i=1,n, for(j=p+1,v[i]-1,vv=concat (vv,j));p=v[i]);return(vv)}
A096032
Take pairs (a, b), sorted on a, such that T(a)+T(b)=concatenation of a and b, where T(k) is the k-th triangular number A000217(k). Sequence gives values of b.
Original entry on oeis.org
1, 415, 1545, 1726, 2196, 910, 3676, 3846, 910, 5226, 415, 6970, 7171, 8526, 9231, 9300, 9756, 9850, 9880, 44835, 9880, 9850, 9756, 9300, 9231, 52830, 8526, 7171, 6970, 5226, 3846, 3676, 2196, 1726, 1545, 84906, 89386, 99580, 99580, 89386, 84906
Offset: 1
1726 of the sequence forms a pair with 150 and we indeed have T(150)+T(1726)=11325+1490401=1501726.
- J. S. Madachy, Madachy's Mathematical Recreations, pp. 166 Dover NY 1979.
-
f[n_] := Block[{k = n + 1, t1 = n(n + 1)/2, td = IntegerDigits[n]}, While[k < 15*n && t1 + k(k + 1)/2 != FromDigits[ Join[ td, IntegerDigits[k]]], k++ ]; If[k != 15*n, k, 0]]; Do[ k = f[n]; If[k != 0, Print[n, " & ", k]], {n, 10^6}] (* Robert G. Wilson v, Jun 21 2004 *)
Comments